JОВАН Д. КЕЧКИЋ
MATEMATIKA SA ZBIRKOM ZADATAKA
za II razred gimnazije
f (x)= xn n ∈ N
f (x)= x 2
f (x)= x 3
f (x)= xn n∈ N
f (x)= 1 x
f (x)= x 2
f (x)= x n (n ∈ N )
xn = a
f (x)= xr (r ∈ Q
= ax 2 + bx + c
2 + bx + c
ax
N N0 Z Q R R+ R+ 0 N = {1, 2, 3,... } N0 = {0}∪ N Z = {..., 3, 2, 1, 0, 1, 2, 3,... }
Q = ß p q p ∈ Z,q ∈ N™ R+ 0 = {0}∪ R+ i N ⊂ N0 ⊂ Z ⊂ Q ⊂ R ii R+ ⊂ R+ 0 ⊂ R iii N ⊂ R+
iv N0 ⊂ R+ 0 v Z ∩ R+ = N vi Z ∩ R+ 0 = N0 vii √2/ ∈ Q √2 ∈ R+ a b ∈ R
(a,b)= {x | a<x<b};[a,b]= {x | a x b}; (a,b]= {x | a<x b};[a,b)= {x | a x<b};
(−∞,a)= {x | x<a};(a, +∞)= {x | x>a};(−∞, +∞)= R. i (1, 3) ∩ (0, 2)=(1, 2)
ii (−∞, 1) ∩ (0, 4)=(0, 1)
iii ( 1, 2] ∪ (2, 5)=( 1, 5)
iv (1, 2) ∩ N = ∅
v [1, 2] ∩ N = {1, 2}
vi (−∞,a) ∪ (a, +∞)= R \{a}
vii (0, +∞)= R+ [0, +∞)= R+ 0
n =
n x + y =6 (0, 6) (1, 5) 1 2 , 11 2 {(t, 6 t) | t ∈ R}
y ∈ R (2) x + y =1,x> 0,y 0. x + y =1 (t, 1 t) t ∈ R x> 0 t> 0 y 0 1 t 0 t 1 2 {(t, 1 t) | 0 <t 1} x y (3) a1 x + b1 y + c1 =0,a2 x + b2 y + c2 =0,
1 2 1 3 0 x ∈ R f (x)
f (x)=3x 1 A = B = R f : R → R x x 2 f (x)= x 2 1 1 1 0 0 1 2 1 2 1 4
f : A → B f (A) A
f (A)= {f (x) | x ∈ A}
f (A)= B f A B f A B y ∈ B x ∈ A
y = f (x)
f (A) f f R R
y ∈ R = B x ∈ R (= A) y =3x 1 y ∈ R x 1 3 (y +1)
f (A)= {f (x) | x ∈ R} = {3x 1 | x ∈ R} = R, R f R R y ∈ R
y< 0 x ∈ R y = x 2
f (R)= {f (x) | x ∈ R} = {x 2 | x ∈ R} = R+ 0 ⊂ R. f (x)= x 2 R R R+ 0
f : A → B x1 x2 ∈ A
f (x1 ) f (x2 ) ∈ B f 1 1 f : A → B 1 1 x1 = x2 ⇒ f (x1 ) = f (x2 ), f (x1 )= f (x2 ) ⇒ x1 = x2 f 1 1 3x1 1=3x2 1 ⇒ x1 = x2 f 1 1 2 =2, ( 2)2 =22 A B R R f x f (x) A f R f (A)
f (x)=3x 1. f R x ∈ R 3x 1 f (R) R
(p ⇒ q ) ⇔ (¬q ⇒¬p)
x ∈ R x 2
f (x)= x 2 R
R f (R) R+ 0 A
f (x)= 1 x
R \{0} x =0 1 x
R f (A)= A = R \{0} x ∈ A f (x)=0 f1 f2 A
⊆ R f1 (x)= f2 (x) x ∈ A
f : A → R A ⊆ R (x,y ) y = f (x) f f (x,f (x)) x ∈ A
y = f (x) f
l l y = f (x) y =
f (x)
f (x)=2x 3 y =2x 3
f (x)=2 3x y =2 3x f x 1 f (x)=1 y =1 x x =2 y y
x1 y1 y2 y3 f (x)= |x| f (x)= x A =(x1 ,y1 ) B =(x2 ,y2 ) |AB |
i A B y x1 = x2
|AB | = y2 y1 , y2 >y1 , = y1 y2 , y2 <y1 ,
|AB | = |y2 y1 |, A B ii A B x y1 = y2
|AB | = |x2 x1 |
n ∈ N
5n 5n +1 +1 : 1 3 · 25n 1 25n = 2 · 5n +1 5n +1 : 1 4 52n 1 (5n )2 = 2 5n +1 5n +1 (1 5n )(1+5n ) (1 2 5n )(1+2 5n ) = 1 5n 1 2 5n . n ∈ N x xn (1) f (x)= xn x xn R R n =1 1
22 · 25 28 :23 (22 )5 2 3 4 43 2 7 3 7 4 3 45 2 7 3 7 4 4
22 · 33 43 10 217 · 39 45 · 67 (3, 5)11 2 3 8 3 7 9
a b c x y ∈ R n k ∈ N Å a 3 b2 c ã5 : Å a 2 b c3 ã2 Å 3a 2 b c ã5 : 3ab 2c 3 Å 2a n bk c15 ã2 : Å 3ab3 c4 ãn (x 2 y 2 ): x y x + y (x 2 y 2 ): x y x2 y 2 (x 2 y 2 )2 :(x + y )2
(x 2 y 2 )2 :(x 3 + y 3 )2 (x 2 y 2 )n :(x + y )n 1 (x 2 y 2 )n : Å x + y x y ãn 2
(1 9x 2 )7 (9x2 6x +1)3 (1 9x 2 )7 (3x 1)5 (1 3x)2n (1 3x)n
(1 3x)2n (3x 1)n (1 3x)n (3x 1)n 2 (a +1)(a 3 1)(a 1) (a2 1)3
a n+2 a n+1 + a n a3 +1 a n+2 bn a n+1 bn+1 + a n bn+2 a3 + b3
a n+2 4a n+1 +4a n
an+2 4an (x 2)2n (2 x)n f (x)= x 2
(1) f (x)= x 2 .
1 x x =0 x x 2 > 0 x =0 x 2 =0 1 R+ 0 1 x =0 (A,B ) f (x) > 0 f f (x) < 0 f
f (x)= x 2 (−∞, 0) (0, +∞) f f (x)=0 f x 2 =0 0
f (x)= x 2 0 m m f (x) x f f m M
f (x) M x f f M
f (x)= x 2 0 x 2 0 x ∈ R M x 2 M x ∈ R y x x x 2 =( x)2 x ∈ R f f (x)= f ( x) x f
f (x)= x 2
f (x)= x n n ∈ N
f 1 1 f ( x)=
f (x) ⇒−x = x (0, 0)
(0, 0) f (x)= x 2 (A,B ) a f (x) f (a)
x ∈ (A,B ) f a
f (a) (A,B ) a f (x) f (a)
f (a)
x ∈ (A,B ) f a
f (x)= x 2 0 0 y = x 2 a b a<b a 2 <b2
f (a) <f (b) 0 <a<b a<b a a 2 <ab b ab<b2 a 2 <b2 (A,B ) f
a<b ⇒ f (a) <f (b), a,b ∈ (A,B ), f (A,B ) f (x)= x 2 (0, +∞)
f (x)= x 2 (0, +∞) y = x 2 a b
a<b a 2 >b2 f (a) >f (b) a<b< 0
a<b a a 2 >ab b ab>b2 a 2 >b2 (A,B ) f
a<b ⇒ f (a) >f (b), a,b ∈ (A,B ), f (A,B )
f (x)= x 2 (−∞, 0)
2 )
(x)= x 2 PQ f (x)= x 2
(A,B ) a b ∈ (A,B )
f (x)= x n n ∈ N
a 2 2ab + b2 , 0 (a b)2 , 3 a b ∈ R
f (x)= x 2 R
f (x)= x 2 R 0, 1 2 y = x 2
f (x)= x 2
f (x)= x 2 1 1 0 x =0 (0, +∞)
(−∞, 0)
(−∞, +∞) f (x)= x 2
(0, 0) f (x)= x 2 R R+ 0 X R+ 0 x R X x X = x 2
f (x)= x 2
f (x)= x 2 (x0 ,x 2 0 ) y = x 2 (x0 , x 2 0 ) y = x 2 (x0 ,x 2 0 ) (x0 , x 2 0 ) x y = x 2 y = x 2 x f (x)= x 2
f (x)= x 2
(
)=
f (x)= x 2
f (x)= x 2 (−∞, 0) (0, +∞)
f (x)= x 2 (0, 0)
(
)= x
(x)= x n n ∈ N
(x)=(x 1)2 (x0 ,y0 ) y = x 2 y0 = x 2 0 y0 =((x0 +1) 1)2 (x0 +1,y0 ) y =(x 1)2 (x0 ,y0 ) y = x 2 (x0 +1,y0 ) y =(x 1)2 (x0 ,y0 )
, 0) y =(x +2)2 y = x 2 2 ( 2, 0)
β< 0 ( α,β ) y = 1 2 (x 3)2 +1 y = 2 Åx + 1 2 ã2 +3
=0
=0 (4) f (x)= a(x + α)2 + β (5) y = a(x + α)2 + β. f (x)= x 2 f (x)= x 2 (0, 0) ( α,β )
i a> 0 5 y = ax 2 4 (−∞, α) ( α, +∞)
ii a< 0 5 y = ax 2 4 (−∞, α) ( α, +∞)
f (x)= x n n ∈ N a =0 ax 2 + bx + c = a x 2 + b a x + c = a Åx 2 +2 b 2a x + b2 4a2 ã b2 4a + c = a x + b 2a 2 + 4ac b2 4a , y = ax 2 + bx + c y = a x +
2 + 4ac b2 4
(a =0) a> 0 a< 0 b 2a , 4ac b2 4a f (x)= ax 2 + bx + c (a =0) R+ f R f (0, +∞) f (−∞, 0)
R f R f (0, +∞) (−∞, 0) f (A,B ) ⊂ R a ∈ (A,B ) f a (A,B ) f A<C<B f (A,C ) (C,B ) C
f [ 5, 5] i ii iii
f (x)= x n n ∈ N
( 3, 27),..., (3, 27) y = x 3
1 1 x> 0 x< 0 0 x> 0 x 3 > 0 x< 0 x 3 < 0 x 3 =0 x =0
(x,x 3 ) ( x, x 3 ) ( x)3 = x 3 f
f ( x)= f (x)( x D x ∈ D ⇒−x ∈ D ) f (x)= x 3 1 1 R a b a 3 b3 =(a b)(a 2 + ab + b2 ) =(a b) ÇÅa + b 2 ã2 + 3 4 b2 å , a + b 2 2 + 3 4 b2 > 0 a 3 b3 a b (a 3 b3 )= (a b). a b< 0 ⇒ a 3 b3 < 0 a<b ⇒ a 3 <b3 f (x)= x 3 R f (x)= x 3 (0, +∞) (−∞, 0) f (a)+ f (b) 2 f Å a + b 2 ã = a3 + b3 2 Å a + b 2 ã3 = 1 8 (4a 3 +4b3 a 3 3a 2 b 3ab2 b3 ) = 3 8 (a 3 + b3 a 2 b ab2 ) = 3 8 ((a + b)(a 2 ab + b2 ) (a + b)ab) = 3 8 (a + b)(a 2 2ab + b2 )= 3 8 (a + b)(a b)2 .
)
3
+ b 2
3 + b3 2 f (x)= x 3 (0, +∞) a 0 b 0 a + b 0 f (x)= x 3 (−∞, 0) y = x 3 y = x 3 f (x)= x 3
R X x
x 3 ( 2, 3) a> 0 f ( a,a) f (0) f R f (0, +∞) f (−∞, 0) f R (0, +∞) f (−∞, 0) R i ii (0, 4) (4, +∞) iii (0, 1) (2, 3) (1, 2) (3, +∞) iv 1 f R (0, +∞) f (−∞, 0) f (x)= x 3
(x)=2x 3
(x)= 1 2 x 3 f (x)= x 3 +1
(x)= x 3 +2
(x)= x 3 1 f (x)=(x 1)3
)=(
(x)=(x +2)3 f (x)= 1 2 (x 3)3 f (x)= 1 3 (x +1)3
(x)= x 3
(1)
(2)
(3)
f (x)= x n n ∈ N
f (x)= 1 2 (x 1)3 +2
f (x)= xn n∈ N
f (x)= xn (n ∈ N). n =2 n =3
f (x)= x 2k (k ∈ N)
f (x)= x 4 f (x)= x 6
f (x)= x 2 2 R+ 0
i x =0 0 ii 0 iii iv (−∞, 0) (0, +∞)
v R
f (x)= x 2 f (x)= x 4 f (x)= x 6
f (x)= x 5 f (x)= x 7
f (x)= x 2k +1 (k ∈ N)
f (x)= x 3 3
i x> 0 x< 0 0
iv R
v (0, +∞) (−∞, 0)
f (x)= x 3 f (x)= x 5 f (x)= x 7 1 n x 0 1 R+ 0 1 R+ 0 R+ 0 X x X = xn
i x> 0 ⇒ xn > 0 ii 0 <a<b ⇒ an <bn ii
f (x)= xn (0, +∞) a<b
an 1 an 2 b an 3 b2 ... abn 2 bn 1
an <an 1 b,an 1 b<an 2 b2 , an 2 b2 <an 3 b3 ,...,a 2 bn 2 <abn 1 ,abn 1 <bn an <an 1 b<an 2 b2 <an 3 b3 < ··· <a2 bn 2 <abn 1 <bn .
(4) a = b ⇔ an = bn (a> 0,b> 0). a = b an = bn an = bn a b a = b an = bn a = b a<b b<a a<b ii an <bn an = bn b<a a = b 4 a> 0 b> 0 a b a< 0 b< 0 a = b ⇔ a n = bn ∧ ab 0 a b 4 n ( 2)4 =24 2 =2 n a<b ⇔ an <bn a b
(5) (p ⇒ q ) ⇔ (¬q ⇒¬p). p b<a q bn <an 5 (6) (b<a ⇒ bn <an ) ⇔ (bn an ⇔ b a),
m n m n 0 m n
(am )n = 1 (am ) n ( ) = 1 am( n) = 1 a mn
( n ∈ N) = amn ( ). m n
(am )n = Å 1 a m ãn ( ) = 1 (a m )n = 1 a mn
( m ∈ N) = amn ( ) m n (am )n = Å 1 a m ãn ( ) = 1 Å 1 a m ã n = 1 1 a( m)( n)
( m, n ∈ N) = 1 1 amn = amn . 3 3 3 4 4 1
. x 4 + x 2 + x 1 x 1 = x 3 + x 3 +1 m · 10n 1 m< 10 n ∈ Z 2, 347=2, 347 1=2, 347 100 , 23, 47=2, 347 · 10=2, 347 · 101 , 234, 7=2, 347 · 100=2, 347 · 102 , 0, 2347=2, 347 1 10 =2, 347 10 1 , 0, 02347=2, 347 1 100 =2, 347 10 2 .
I = 0, 002304 · 0, 0007 0, 0013 . 0, 002304=2, 304 × 10 3 0, 0007=7 × 10 4 0, 0013=1, 3 × 10 3 I = 2, 304 10 3 7 10 4 1, 3 10 3 = 2, 304 7 1, 3 10 3 10 4 10 3 ≈ 12, 406154 · 10 4 =0, 0012406154.
=0
n
\{0} R a =0 a 0 =1 a =0 n ∈ N a n = 1 an a n a =0 n m n a =0 b =0
a m a n = a m+n a m an = a m n (a m )n = a mn (ab)n = a n bn Ä a b än = a n bn 20 +2 1 +2 2 +2 3
( 3)2 ( 2) 2 +( 1) 3 05 ( 3)3 ( 3) 2 :6 2 3 3 ( 2) 3 6 2 3 2 812 272 ( 3) 3 2 4 +3 2 52 · 32 1 Å 2 3 1 3 2 2 ã 1 75 0, 15 1 0, 002 125 · ( 5) 4 · Ä625 1 · 53 ä 1
8 · 49 2 ·
· 125 · 1 5 1 354 7 2 a b c x =0 n k ∈ N a 2 a 7 a3 a 3n a 4n a n 4a k a k +1 (2a 2k ) 1 (3a 5 : a 2 )3 (3a2 : a 3 )2 (2a 2 b) 3 8a 5 b4 1 2 a 3 b2 c 2 Å2 a 7 b 3 b2 c4 ã 1 (52 a n b3 c 2k )3
25an b 2 c 3k )2 a 2 +2(a + b)2 4ab(a 2 + b2 )0
(a + ab 1 )2 b + Ä a b ä 1 2
(x 2 a 2 )(a 1 x 1 ) 1
(a 2 x 2 )(x 1 + a 1 ) 1
(x 4 a 4 )(x 2 a 2 ) 1
a 2 + b 2 2
(a4 b 4 )2 = Äa 2 b 2 ä 2 a 6 b 6 a 3 + b 3 = a 3 b 3
Ä4a 2 b2 ä2n : 1+ 8ab 4a2 4ab + b2 n =(2a + b)4n
2 3 a 2 b 4 c 5 (bc =0) Å 9+ x 2 9 x 2 ã 1 x =0; x 2 = 1 9
Å 34 a 1 b6 : b 8 3 5 a 7 ã 2 (ab =0) 8x 2 1 x (4x) 1 x (x =0
3+ a 5a 7x 3 y 2 3x7 y 5 2 3 x 3 y 3 (8xy 2 ) 4 a + b ab
a 4 + a 2 b2 + b4 a4 b4 1+2a +4a 2 2a(1+2a) x 1 27a 3 x 4 x 1 +3ax 2 +9a2 x 3
3a 2 b 7
(9 1 a3 b 4 ) 2 Å a 1 b 2 x3 y 3 ã 2 : Å ab 3 xy 2 ã 3
Å 2a 2 b3 3a4 b ã 2 Å 4a 1 b3 9a5 b 2 ã Å 5x 3 y 2 3xy 4 ã : Å 25x 3 y 1 9x 2 y 5 ã 1
7a n+1 bn 1 4x2 n y 2+n : 7a n 1 b1+n 3x1 n y 1+n a 2 b 1 2 a 3 b2 a7 (cb2 ) 1 a 12 b3 a 2 c4 : a 2 b5 c 3
1+ x 2 1 x 2 1 2x 1 x 1+ x 2 1 x 1 1 2x 1 x
1+ x 1 1 x 2 1 2x 1 x a 2 b 2 (b 1 a 1 ) 1
Äa 2 b 2 ä 1 Äb 1 a 1 ä 1 x 3 + y 3 x 2 x 1 y 1 + y 2
Äx 6 y 6 äÄx 4 + x 2 y 2 + y 4 ä 1
Äx 2 a 2 äÄa 1 + x 1 ä 1
Äx 6 a 6 äÄa 3 + x 3 ä 1
Äx 4 a 4 ä 1 Äx 2 a 2 ä
Ä(a x) 1 (a + x) 1 äÄ(a x) 1 +(a + x) 1 ä 1
Å a 2 x 2 x 2 a 2 ã Ä a x ä 1 + Ä x a ä 1 1
Å a + a 1 b2
a a 1 b2 1ã : 1 a b + 1 a + b
a 2 b3 a a 1 b2 a : Å1+ ab 1 a 1 b +1 ã
b b 2
b b 1 : b +1+ b 1 a a n · 10m m,n ∈ Z
30005 9007 800003 :2005 0, 000276 :0, 003 2
0, 0042 0, 00065 0, 0243 ( 0, 01)5 ( 0, 002) 3 0, 08 0, 0016 2
b =0, 0003 · 0, 004 a
b = a 102 b = a 10 1
b = a · 10 6 b = a · 10 10 x =125000 · 0, 021 a x = a · 10 3 x = a · 10 1 x = a · 100 x = a 102 x = a 104 ab = Äa 4 b 4 äÄa 2 b 2 ä 1 = b 6 a 6 a 3 + b 3 1 b 3 a 3 a b f (x)= 1 x (1) f (x)= 1 x
(1) (1) 1 x = 1 x , f ( x)= f (x) x x> 0 (1) x 1 x =0 x
R\{0}
= 1 x x y y = 1 x y = 1 x x
(x)= 1 x
0
(x)= a x
0 f (x)= 1 x
(x)= 1 x2 R\{0} R+ (1) (1) (1)
(1) (1) (0, +∞)
(−∞, 0) (1) (0, +∞)
(−∞, 0) x x 2 x x x x 2 y x
(1) R\{0} R+
f (x)= 1 (x 1)2 +2 f (x)= 1 (x +2)2 1
f (x)= x n (n ∈ N ) n
f (x)= 1 xn (1)
f (x)= x 2 n
f (x)= x 1 (1) R+ n ∈ N
(1) n =1 n =2 n =3 R+
f (x)= x n (n ∈ N)
n ∈ N f (x)= x n
(−∞, 0) (0, +∞)
f (x)= x 1 R\{0}
f (x)= x n (n ∈ N)
f (x)= 1 x +3 1 f (x)= 2x +1 x 2 f (x)= x +2 2x 1
f (x)= 1 x2 1 f (x)= 1 x2 +2
(x)= 1 (x 1)2 f (x)= 1 (x +2)2
f (x)= xn x 0 n ∈ N
f (x)= xn a n b bn = a f (x)= xn x 0 y = xn a
= x n y = x n a> 0 n ∈ N b> 0 bn = a y = x n a b1 b2 b3 b n 1 = a
b n 2 = a b n 3 = a y = x n y = M a M b bn = a
= x n
0 b> 0 bn = a f (x)= x n
(2)
(3)
, 2)
= √x a> 0 (a,b) y = x 2 (b,a) y = √x (a,b) y = x 2 b = a 2 a> 0 b = a 2 ⇔
,
= √b (b,a) y = √x (a,b) (b,a) y = x y = x 2 (x 0) y = √x
f (x)= n √x n n f (x)= n √x x 0 f (x)= xn y = x
n
f (x)= n √x x ∈ R
f (x)= xn y = x
a<b ⇒ n √a< n √b n a> 0 b> 0 n a,b ∈ R
f (x)= n √x
n ∈ N
f (x)= n √x R
n ∈ N f (x)= n √x R+ 0
f (x)= n √x n ∈ N
n f (x)= n √x
n f (x)= n √x
f (x)= 3 √x f (x)= x 3
f (x)= √x f (x)= x 2 (x 0) x 0
n a b
a<b ⇔ n √a< n √b
f (x)= √x +1
(x)= 3 √x 1 f (x)= √x 1 f (x)= 3 √x +1 f (x)= √x 1 2 f (x)= 3 √x +1+2 f (x)= √x f (x)= |x|
x 2 5=0 5x 3 +80=0 5x 4 +80=0
75x 2 3=0 125x 3 1=0 75 3x 4 =0 3x 2 a =0 x 3 27a =0 x 4 9a 4 =0 (x +5)2 =125 (x 4)2 529=0 (x +3)3 +64=0 (x +3)4 +64=0 (5x +4)2 a 2 =0 (5x +4)2 + a 2 =0 (5x +4)2 + a =0 (2x +1)3 + a =0
(6) (5)
m n
m n = n √a
a m n n m = a a f (x)= xr (r ∈ Q r = m n m ∈ Z n ∈ N m n f (x)= xr r = m n (1) r =0 r =1 n (1) R+ 0 r> 0 R+ r< 0 (1) R+ r> 0 f (0)=0 r> 0 (1) (0, +∞) r< 0 (0, +∞)
r> 1 r< 0 (1) (0, +∞)
0 <r< 1 (1) (0, +∞)
r> 1
0 <r< 1 r< 0
n m
(1) R r> 0 R\{0} r< 0 (1) x =0 r> 0 f (0)=0 (1)
r> 0 (1) (0, +∞) (−∞, 0) r< 0 (1) (−∞, 0) (0, +∞) r> 1 (1) R r< 0 (1) (−∞, 0) (0, +∞) 0 <r< 1 (1) (−∞, 0) (0, +∞) n m (1) R r> 0 R\{0} r< 0 (1) (0, +∞) (−∞, 0) r> 0 f (0)=0 (1)
r> 0 (1) R r< 0 (1) (−∞, 0) (0, +∞) r> 1 r< 0 (1) (0, +∞) (−∞, 0) 0 <r< 1 (1) (−∞, 0) (0, +∞)
A> 0 B> 0 (3) a b> 0 (2)
(0, +∞) (−∞, 0)
(2)
m n m ∈ Z n ∈ N f (x)= x m n x> 0 x 0 x =0 x ∈ R
(0, +∞ (0, +∞) R (0, +∞) (−∞, 0) (−∞, 0) (0, +∞) (−∞, 0) (0, +∞)
r ∈ N a ∈ R
r ∈ Z a ∈ R\{0}
r ∈ Q a ∈ R+ ar a> 0 r ar r ∈ R\Q a> 1 a r> 0 r ∈ Q ar > 1
r = p q > 0 p q a> 1
ap > 1 q √ap > 1 q √ap 1 q ap 1 ap > 1 b x1 x2 x1 <x2 ax1 <ax2 x1 <x2 r (= x2 x1 ) x1 + r = x2 ax2 = ax1 +r = ax1 ar >ax1 a
ar > 1 a√2 a> 1 a√2
a,b ∈ R+ r,s ∈ R
ar as = ar +s , ar as = ar s , (ar )s = ars , (ab)r = ar br , a b r = ar br √2 √3 3 √2 n √k n k
n k x n = k
n √k n √k n √k ∈ Q n √k / ∈ Q n k x n = k (1) k n √k n √k k p q p,q ∈ N (1) p q p q (2) p n q n = k q n 1 (3) p n q = kq n 1
(6)
(7)
(9)
+ p> 0 (x + p)2 <a
+ p ∈ S
+ p>x x S x 2 >a x 2 a 2x +1 > 0 q 0 <q< x 2 a 2x +1 . (x q )2 = x 2 2xq + q 2 >x2 2xq q q> 0 ⇒ q 2 > q = x 2 q (2x +1) >x2 (x 2 a) Å q< x 2 a 2x +1 ⇔−q (2x +1) > (x 2 a)ã
q S x q<x x S
2 <a x 2 >a x 2 = a
2 = a √a x a> 0 √a a n x x n = a
r a> 0 r ∈ R a r a> 0 r ∈ Q a r a> 1 0 <a< 1 a r = 1 a r 1 a > 1 a =1 a r =1 a> 1
r = {ap | p<r,p ∈ Q}. r
1 r2 r1 <r<r2
= {ap | p<r,p ∈ Q} a r1 ∈ A (b)
r2
a r a r ap p<r p ∈ Q
0 b2 4ac> 0
x 4ac b2 4a < 0
x
<α<β a> 0 b2 4ac =0 x T = b 2a , 0 ab> 0 a> 0 b2 4ac =0 x x 4ac b2 4a > 0 ab< 0 a< 0 b2 4ac> 0 (2) α β x α β x 4ac b2 4a > 0 α< 0 β> 0
a< 0 b2 4ac =0 x T = b 2a , 0
ab> 0
a< 0 b2 4ac< 0 x x 4ac b2 4a < 0
ab< 0 (1) a T =
b 2a , 4ac b2 4a ã . x (2)
y (0,c) y =2x 2 +3x +1 3 4 , 1 8 x 2x 2 +3x +1=0 1 1 2 (0, 1)
=3x 2 +2x +1
3 , 2 3 x 3x 2 +2x +1=0 (0, 1) y = 4x 2 +4x 1 1 2 , 0 (0, 1) y = ax 2 + bx + c (a,b,c ∈ R; a =0) (3) (3) (3) (3) x x (3) x (3) x (3) x a
= 3 2 (x 2 +2)+6 y = x 2 +4x +3 y = x 2 2x +4 y = x 2 2x y =4x 2 4x 1 y =2x 2 +4x 1 y = 3x 2 +12x 10 y = 1 2 x 2 3x +3 y = x 2 +4x +5 y = 2(x +5)(x 1) y = x 2 + kx + k +1 k ∈ R y = x 2 2(a 2)x +2(a 2) a ∈ R y = x 2 (3 k )x +2k 1 k ∈ R f (x)+ ax 2 + bx + c (a =0) (1) a b c x ax 2 + bx + c (1) R y = ax 2 + bx + c (1) a> 0 a< 0 k y = x 2 +kx+k +1
y = ax 2 + bx + c x (3)
y = ax 2 + bx + c x (3)
f (x)=2x 2 3x +5 f (x)=5x 2 2x 7 f (x)= 3x 2 +4x 5
f (x)= x 2 +13x 4 f (x)= 1 4 x 2 1 2 x + 3 4
f (x)=2x 2 1
f (x)=2x 2 +4x 3
f (x)= 3x 2 +5x f (x)= 1 2 x 2 +11x 2
f (x)= x 2 3x f (x)=3x 2 +5x +8
f (x)= 2 3 x 2 +4x 9 f (x)= x 2 17x 1
f (x)= x 2 x 6 f (x)= 1 2 x 2 3x + 5 2
f (x)=3x 2 4x +12 f (x)= x 2 + 7 2 x +2
f (x)=5x 2 3x +4
f (x)=4x 2 12x +9
f (x)= x 2 +4x f (x)= 16x 2 +48x +85
f (x)= x 2 4 f (x)= x 2 4x f (x)= x 2 4|x|
f (x)= |x 2 4x| f (x)= |1 x 2 | f (x)= x 2 4x 5
f (x)= 2x 2 +5x +3 f (x)= 1 3 x 2 +2x 16 3 f (x)= 3x 2 + 9 2 x +3 a b c a =0 x ax 2 + bx + c> 0,ax 2 + bx + c 0, ax 2 + bx + c< 0,ax 2 + bx + c 0 (1) (1) x (1)
f (x)= ax 2 + bx + c
y = ax 2 + bx + c; (2)
+ b
+ d > 0 (ax + b)(cx + d) > 0
(5) (6)
(16) x 2 +2x 11 2(x 3) ? x 2 +2x 11 2(x 3) = y,
)= x 2 +2x 11 2(x 3) (−∞, 2] ∪ [6, +∞)
(2x +1)(x 3) x 5 > 0
x 2)(5 x) x2 +4
x 2)(5 x) x2 4
x 2 + x> 1 5x x 2 > 7 24x 9x 2 16 x 3 >x x 4 x x 5 <x 3x(x 4) x 2 < 5 x 2 7x +6 x2 1 <
x
x x 2 2x2 9x +7 < 1 2x +5 x +1 2 x x 2 50 2x2 +7 7 x 2 x2 +3x 4 > 1 3 3 < x 2 + x 2 x2 x +1 < 2 |x 2 4| > 3 |x 2 1| < 1 x 2 4|x|− 5 >
R (
+1)x
∈ R
+1) >
∈ R x (k 1)x 2 +(k 5)x (k +2)=0 m 9 m2 16 2x m(m +4) x 2 m(m 4) =0 k x 2x x +3 + kx 2 x2 9 = x 1 x 3 x ∈ R 1 11 x x2 5x +9 1 x ∈ R 1 3 x 2 x +1 x2 + x +1 3 x ∈ R x 2 +34x 71 x2 +2x 7 / ∈ (5, 9) x +2 2x2 +3x +6 a<b x ∈ R (x + b)2 4ab 2(x a) / ∈ (2a, 2b) √2 x = x (2 x)2 = x 3 (2 x)3 = x |2 x| = x 2 x = x
(3) (4) (5)
(14) (14) (13)
(15) (13) (15)
(22) (21) (20) (22) (20)
(23) (23) (22) (20) (20)
3x +8= x +2 x 2 +3 x2 +3=2
2x 2 + x + 2x2 + x +1=5 x 2 +6x +2 x2 +6x =24
3x 2 16x +3 3x2 16x +21=7 3x 2 4x + 3x2 4x 6=18 8+9 (3x 1)(x 2)=3x 2 7x 3 √2x √2x 5 = 59 10 5… 3 x +7 x 3 = 68 3 … x 1 x + … 1 x x = 13 6
x +3 2x 1 + … 2x 1 x +3 = 5 2 2x2 +5x 2+ 2x2 +5x 9=7 x 2+ √2x 5+ x +2+3√2x 5=7√2
2x2 +5x 2 2x2 +5x 9=1
3x2 2x +9+ 3x2 2x 4=13
2x2 7x +1 2x2 9x +4=1 x2 +6x +6+ x2 3x +3=3 x2 +5x +2 x2 3x +3=3
3x2 7x 30 2x2 7x 5= x 5 4x2 7x 15 x2 3x = x2 9
2x2 9x +4+3√2x 1= 2x2 +21x 11
2x2 +5x 7+ 3(x 6)(x 1)= 7x2 6x 1
8√x 5 3x 7 = √3x 7 x 5 18(7x 3) 2x +1 = 250√2x +1 3√7x 3
(4) x< 7 9 ∧ x 3, 3, 7 9 x 2 3x +2 0 (−∞, 1] [2, ∞) 3+ x< 0 (−∞, 3) (5) (−∞, 3) (3) 3, 7 9 (−∞, 3) −∞, 7 9 x
x 1 > 2
2x 3 > 1
x +2 < 5 3+ x2 2
x 3 2x +1 < 4 2x2 3x +1 1 x2 4x +2 < 2 3 2x x2 > 1 √x 2+ √3 x> 1 √x +5+ √2 x> 3 (x 3)(x +5) < 7 x 3x2 5x 3 > √2x +3 1 √1 4x2 x < 3(x =0)
ad2 + bd + c =0 4a 2 d2 +4abd +4ac =0
2
+
(ai)2 = a 2 i2 = a 2 , (a + bi)i = ai + bi2 = ai b i2 = 1 (a + bi)c = ac + bci,i3 = i2 i =( 1)i = i, a + bi i i2 1 i x x 2 0 i i2 = 1 ( i)2 = 1 ( i)2 =( 1)
i
, (a i)(a + i)=0, a = i a = i x x 2 = 1 i i i a + bi a b
C = {a + bi|a,b ∈ R.} b =0 a + bi a
a =0 a + bi bi
z = a + bi (a,b ∈ R) a b a = z,b = z z = z + i z. (2+3i)=2 ( 1 7i)= 7 5=5 5=0 i =0 i =1
a + bi a,b ∈ R i2 = 1 i i2
a,b ∈ R
a + bi =0 ⇔ a =0 ∧ b =0. a =0 ∧ b =0 a + bi =0 a + bi =0 (1) b =0 (1) i = a b a b i b =0 (1) a =0 a,b,c,d ∈ R a + bi = c + di ⇔ a = c ∧ b = d. a + bi = c + di ⇔ (a c)+(b d)i =0 ⇔ a c =0 ∧ b d =0 ⇔ a = c ∧ b = d.
a + bi c + di A + Bi A,B ∈ R
a + bi c + di a,b,c,d ∈ R (a + bi)+(c + di)=(a + c)+(b + d)i, (a + bi) (c + di)=(a c)+(b d)i. a + c b + d a c b d ∈ R (a + bi)(c + di)= ac + adi + bci + bdi2 =(ac bd)+(ad + bc)i i2 1 ac bd ad + bc ∈ R a + bi c + di c + di =0 c + di =0 i c d a + bi c + di = a + bi c + di c di c di = ac adi + bci bdi2 c2 cdi + cdi d2 i2
+ bd c2 + d2 bc ad c2 + d2 ∈ R
z a
b z = a + bi (a,b ∈ R) a bi a + bi z
z = a + bi ⇔ z = a bi. z =1+2i z =1 2i 1+2i =1 2i
2 i =2+ i 5=5 i = i 3+5i = 3 5i
z w i z =0 ⇔ z =0
z = w ⇔ z = w
z = z
z + w = z + w
z w = z w
zw = z · w
z w = z w (w =0)
z z ∈ R z z 0
z = 1 2 (z + z ) = 1 2i (z z )
x z = z ⇔ z ∈ R x
z = a + bi w = c + di
a,b,c,d ∈ R
zw =(a + bi)(c + di)=(ac bd)+(ad + bc)i,
zw =(ac bd) (ad + bc)i.
z = a bi w = c di
z w =(a bi)(c di)=(ac bd) (ad + bc)i.
zw = z w
z = a + bi a,b ∈ R z = z bi z = z ⇔ a + bi = a bi ⇔ b = b ⇔ b =0 ⇔ z ∈ R, (x) z z z z z z a b c z
az 2 + bz + c =0, (3)
az 2 + bz + c =0 (4) z =
x + iy (x,y ∈ R) (3)
a(x + iy )2 + b(x + iy )+ c =0,
ax 2 ay 2 + bx + c + i(2axy + by )=0,
C
z ∈ C z = z z
a,b,c ∈ R az 2 + bz + c =0 ⇔ az 2 + bz + c =0 a,b,c ∈ C z w
i z = w ⇒|z | = |w | z = w ⇔|z | = |w | |z + w | |z | + |w | z,w ∈ C
z w z1 =3 2i z2 =2+ i z1 +2z2 z1 z 2 z 1 (z2 3) |z1 z2 | (z1 2i): z2
z = 5 i 3+2i z |z |
z = 3+4i i17 (1 2i)2 z |z | (3 i)2 (2+3i)(4 i) 1 i 5+4i 3 i (1 i)2 1 2i +(1+ i)3 A + Bi A,B ∈ R 3i
2+2i 2i 1 i 1 i 8(19 4i) 5+ i (2 i)3 3+ i (1 i)10 6i (1 2i)3
2+4i 3 i (2+ i)2 2i 3i15 +(2i +2)3 (3 i)2 2+ i
√3 i√2
2√3 i√2 (a + ib)2 a ib (a ib)2 a + ib (a 2 + b2 > 0) x y 2x i +3i(y +2i)=4+5i x y (2i 5)x +(3 i)(x + iy )=(3 2i)2 x y (x +2i)i (x y )(3 4i)= (5 3i)(2+ i) z |z |− z =3+2i z |z 1|− 2z +4i =0 w =4+3i z zw =29 z w = 3 25 ω = 1 2 + i √3 2 (1+ ω 2 )4 = ω (1 ω + ω 2 )(1+ ω ω 2 )=4 (1 ω )(1 ω 2 )(1 ω 4 )(1 ω 5 )=9 (2+5ω +2ω 2 )6 =(2+2ω +5ω 2 )6 =729 (x + y + z )(x + yω + zω 2 )(x + yω 2 + zω )= x 3 + y 3 + z 3 3xyz
4 5 3 7
± 5i
i(a b) a,b ∈ R 2√5 1 √2
x 2 +13x 12
2 +3x +3
x 2 x 1
x 2 +28x +49
b3 = ac(3b a c)
2 + bx + c =0 2b2 =9ac
β x 2 +5x +10=0
α 3β 2β 3α
β
2 + bx + c =0
2 14x +29
b c
=1+2√3
β =1 2√3
±
2 + bp + p =0
= (p +1)
p 2 bp + p =0 b = p +1 (9) b = ±(p +1) b2 4p b = ±(p +1) p p =12 b2 4p = b2 48 b = ±7 b = ±8 b = ±13 x 2 + bx +12=0. (10) ±1 ±2 ±3 ±4 ±6 ±12 (10) x =1 1+ b +12=0, b = 13, x = 11 b +12=0,b =13, x =2 4+2b +12=0,b = 8, x = 24 2b +12=0,b =8, x =3 9+3b +12=0,b = 7, x = 39 3b +12=0,b =7, x =4 16+4b +12=0,b = 7, x = 416 4b +12=0,b =7, x =6 36+6b +12=0,b = 8, x = 636 6b +12=0,b =8, x =12 144+12b +12=0,b = 13, x = 12 144 12b +12=0,b =13. b2 48 b = ±7 ±8 ±13 A + B √C ax 2 + bx + c =0 a b c A B C ∈ Q
∈ Z
p q ax 2 + bx + c =0
∈ Z
2 + bx + c =0
x 2 +3x +4=0 (15) 1 2 (3 ± i√7) (14) a b c d ax 3 + bx2 + bx + a =0
3 bx2 + bx a =0 1 x 4 5x 3 +7x 2 x 2=0 (16) 1+ √2 (16) 1+ √2
x 2
2 2x 1=0, x 4 5x 3 +7x 2 x 2 x 2 2x 1 (x 4 +5x 3 7x 2 x 2):(x 2 2x 1)= x 2 3x +2 x 4 2x 3 x 2 3x 3 6x 2 x 3x 3 +6x 2 +3x 2x 2 4x 2 2x 2 4x 2 (16) (x 2 2x 1)(x 2 3x +2)=0, x 2 3x +2=0 (16) 1+ √2 1 √2 x +3 x 3 + x 3 x +3 =3
+ x +
a ∈ R) 2x x +3 + 8x 2 x2 9 = x 1 x 3 x +1 x 1 2 4 x 1 x +1 2 =4 x 2 2x 1 8=0 x 3 2 x 1 2 =0 x 7 3 x 1 2 =0 x 3 x 2 6x =0 x 5 2 +2x 3 2 8x 1 2 =0 x 7 3 +7x 4 3 8x 1 3 =0
+2x
(10)
(11)
(12)
(13)
(13)
x
8x 3 2n 8x 3 2n =63(n ∈ N) 3x 1 2n x 1 n 2=0(n ∈ N) 2 x 2 +3… 2 x = 11 2 5… 3 x +7 x 3 = 68 3
x 1 x + … 1 x x = 13 6 x 4 7=0
x 4 5x 2 +6=0 x 6 7x 3 8=0
x 2 50 2x2 +7 =7 21 x2 4x +10 = x 2 4x +6
x 4 + x 3 4x 2 + x +1=0 6x 4 25x 3 +12x 2 +25x +6=0 (x 1)(x 2)(x 3)(x 4)=24 (x 7)(x 3)(x +5)(x +1)=1680
(2x 7)(x 2 9)(2x +5)=91 x(2x +1)(x 2)(2x 3)=63 R
i, 1+ i) (1 2i, 1 i) (1)
x y
2x y =0 x 2 2x + y 2 =0 x y +7=0 x 2 +3xy y 2 +2x 5y +64=0
x + y =4 x 2 2y 2 xy +5x +15=0 xy =5 x 2y =3
3x 1 x +1 + 1 y 1 = y +2 (y 1)(x +1) 4y x =7 1 x +2y + y =2 y x +2y = 3 1 x y + x =1 x x y = 2 x y =3 x 2 + y 2 =4 x 2 +2y =10 x y = 1 x 2 y =8 x y =2 4s a x y
ax 2 + bxy + cy 2 + dx + ey + f =0 ∧ Ax2 + Bxy + Cy 2 + Dx + Ey + F =0 a b c d e f A B C D E F a b c A B C
(8)
(8) (t,t) (2t,t) t
(11) (10)
(14) (13) (14) (13)
(17) (16) (17) (16) (16)
(23)
(25)
(26)
(28)
(29)
(25) x y ax 2 + by 2 = c ∧ Ax2 + By 2 = C a,b,c,A,B,C ∈ R x y ax 2 + bxy + cy 2 =
0 ∧ Ax2 + Bxy + Cy 2 + Dx + Ey + F =0 a,b,c,A,B,C,D,E,F ∈ R x y
x 2 + y 2 =5 4x 2 +9y 2 =25
5x 2 6y 2 =111 7x 2 +3y 2 =714
2x 2 2xy y 2 =39 x 2 +2xy +4y 2 =39
x 2 2xy y 2 =2 xy + y 2 =4
9x 2 24xy +16y 2 =54 3(3x 4y ) x 2 + xy 2y 2 =0
x y y x = 9 20 x 2 y 2 =9
x 2 + y 2 + x + y =8 x 2 + y 2 + xy =7
x 2 + xy + y 2 =4 x + xy + y =2
x 2 + y 2 + x 3y 2=0 x 2 + y 2 5x y 2=0
x 2 +4y 2 15x =10(3y 8) xy =6
9x 2 +33x 12=12xy 4y 2 +22y x 2 xy =18
2x 2 xy + y 2 =2y 2x 2 +4xy =5y
x(x 1)= y (y +2) (x + y )2 + x +2y =0
x(a + x)= y (b + y ) ax + by =(x + y )2 ab> 0 a = b
x + 1 y = a y + 1 x = b (a,b ∈ R)
x 2 + y 2 x = a x 2 + y 2 y = b (a,b ∈ R)
x 2 xy =2(a +1) y 2 xy =2(1 a)(a ∈ R) h P h
(4)
0 B ∈ R AB
∈ Q
∈ R+ B ∈ R xa
∈ R\Q
(x)= xa (1) R+ a
∈ N (1) R
∈ Z R\{0} (1) ax a> 0
∈ R a f (x)= ax R f (x)= x 2 f (x)=2x
(x)= x a
(x)= a x f (x)=2x (1)
X> 0 x ∈ R
X =2x x (1) f (x)=3x f (x)= Å 3 2 ãx
f (x)= ax a> 1 y =2x y =3x y = Å 3 2 ãx 0 <a< 1 f (x)= ax a = 1 2 x 3 2
a = 1 3 0 <a< 1 f (x)= ax
i ax > 0 x ∈ R x> 0 x < 1 x< 0 ax > 1 R 1 1 R R+ R f (x)= ax a> 1
0 <a< 1 x1 x2 x1 <x2 r x1 <r<x2
r1 r2 x r1 < x<r2 ar1 <ax <ar2 (a> 1)
f (x)= ax (a> 1). (1)
i x ∈ R ax > 0
M x ax >M (1) R R+ M (> 0)
b a b = M n n>b a n >ab = M x a x m x 0 <ax < m m a x >m x ∈ R a n >m n ∈ N 1 an >m, a n < 1 m , n a n < 1 m x a x (1) a> 1
(x)= ax (0 <a< 1) (5) a = 1 b
(x)= 1 bx b> 1 (5) i (1) a f (x)= a x R R R R
a ∈ R+
x> 0 ⇒ a x < 1 x< 0 ⇒ a x > 1
x> 0 ⇒ a x > 1 x< 0 ⇒ a x < 1
m ∈ R+ ⇒ x ∈ R+ 0 <ax <m
M ∈ R+ ⇒ x ∈ R+ a x >M f (x)=2x
f (x)=2x +1 f (x)=2x + 3 2
f (x)=2x 1 f (x)=2x 3
f (x)=3x
f (x)=3x 1 f (x)=3x 2
f (x)=3x+1 f (x)= 1 3 x+1
f (x)=3x 1 +1 f (x)=2x+1 1 f (x)=3 x f (x)=31 x
2 1 8 = 3 2 ( 8)
0 <a< 1 (4) (5) f (x)= 2 x f (x)=2x (a,b) y =2x b =2a a = 2 b (b,a) y = 2 x (a,b) (b,a) y = x y =2x y = 2 x y = x a> 0 a =1 f (x)= a x f (x)= a x a f (x)= a x (1, +∞) (0, 1) R+ R+ R+ R+ a ∈ R+ x> 1 ⇒ a x> 0 0 <x< 1 ⇒ a x< 0 x> 1 ⇒ a x< 0 0 <x< 1 ⇒ a x> 0 f f (x)= 10 x f (x)= 1 4 x f (x)= 3 x 1 9 , 2 f f (x)= 2 (x 1) f (x)= 3 (x +1) f (x)= 3 (x +1) f (x)=3 3(x +1) f (x)= 2 (2 x) f (x)= 1 2 |x|
f (x)= 2 x + | 2 x| f (x)= 2 x 2 | 2 x|
a xy = a x + a y ; a x y = a x a y, a xr = r a x, 10 x x x (1) x = 3521 · (0, 23)1,7 8, 345 . (1) x = 3521 (0, 23)1,7 8, 345 = 1 2 3521 · (0, 23)1,7 8, 345 = 1 2 Ä 3521+ (0, 23)1,7 8, 345ä , (2) x = 1 2 ( 3521+1, 7 0, 23 8, 345) . 3521 0, 23 8, 345 (2) (1) x x 3521=3, 54667; 0, 23= 0, 63827; 8, 345=0, 92143, x = 1 2 3, 54667+(1, 7)( 0, 63827) 0, 92143 = 1 2 (3, 54667 1, 08506 0, 92143)= 1 2 (1, 54018)=0, 77009. (3) x x =5, 8896
d1 d2 d3 d4 d5
374=57287 t 3, 74 10k
d1 d2 d3 d4 d5 d1 d2 d3 d4 d5
3, 74=0, 57287 37, 4=1, 57287 374=2, 57287 3740= 3, 57287 0, 374=0, 57287 1= 0, 42713 0, 0374=0, 57287 2= 1, 42713
= k +0,d1 d2 d3 d4 d5 k k d1 d2 d3 d4 d5 ... x · 10k (1 x< 10) x =0 d1 d2 d3 d4 d5 ... k 1 x< 10 1, 001, 1, 002, 1, 003,..., 9, 997, 9, 998, 9, 999. 3, 521=0, 54667 k 35, 21=1, 54667, 35210=4, 54667, 0, 03521= 2+0, 54667= 1, 45333, 35214
35210 < 35214 < 35220, 4, 54667 < 35214 < 4, 54679,
35214=4, 54667+ p, p 0 <p< 0, 00012 p p =0, 000048
35214=4, 54667+0, 000048=4, 546718,
35214=4, 54672. 1 x< 10 35214 x x x =3, 54617
(8) (7)
(10) (10) t i 10 x =1 10 y =2 ⇔ x =10 y =100 10 x =2 10 y =1 ⇔ x =100 y =10 (9) x y (10, 100) (100, 10) x =2x , 3x =2x (11) 2 x =2 x (12)
(3x 5)=0
(2x x 2 )=0 10 (x 2 x)=0 10 ( 2 3x)=2
(3x +
x
x 7 (2x 1)+ 7 (2x 7)=1
(4+2x+2 )=2 10 2+ 10 (5 24 x 1)
(3x +2) 10 (2 · 32 x +9)= 10 3
(9x 1 +7)=2+ 2 (3x 1 +1)
2 10 3 =8x +5
1 2x =3 1 2 x 52 x =2 1 2 x 3 x 10 x = 100 x x1+ 10 x = x x 100 x 5 √x = 5 40 1 2 (x 2 +2x 7)= 1 2 2 (x 2 6x +9)
x = 1 3 (x +1) 2 x = 8 x 1 3 x = 1 3 (1 x) 1 3 x =3x
3 (x 2 5x +6) < 0 0,5 (x 2 +1) 0,5 (5x 3) 1 2 x 1 2 x 1 < 1 x 2 2 2 x 2 x 2 > 1 x x> 0 A B R f 1 1 A B i X ∈ B x ∈ A f (x)= X f x ∈ A f (x)= X f 1 1 f 1 1 A B X ∈ B x ∈ A f (x)= X f A → B 1 1 f 1 f 1 (X )= x (X ∈ B ) x A f (x)= X f 1 (X )= x ⇔ f (x)= X (x ∈ A,X ∈ B ). (1) f 1 f f x ∈ A X ∈ B f 1 X ∈ B x ∈ A x ∈ A (2) f 1 (f (x))= x; X ∈ B (3) f f 1 (X ) = X ; X f (x) x f 1 (X ) f 1 f
(1) x ∈ A f (x)= X (X ∈ B ). (4)
(4) X ∈ B x f 1 (X ) f 1 (1) f 1 B → A f A → B f f 1 f f 1 f (x)=2x 3 1 1 R R f 1 f (x)= X x ∈ R x ∈ R 2x 3= X (x ∈ R) x = 1 2 (X +3), f 1 (X )= 1 2 (X +3) (5)
1 f (x)=2x 3 (6) f 1 (x)= 1 2 (x +3). (2) (3) f 1 f (x) = 1 2 f (x)+3 = 1 2 (2x 3)+3 = x; f f 1 (x) =2f 1 (x) 3=2 1 2 (x +3) 3= x. (5) (6) f 1 (t)= 1 2 (t +3)(t ∈ R) f 1 ( )= 1 2 ( +3)( ∈ R). X x t f (x)= x 3 1 1 R R f 1 R → R x ∈ R x 3 = X (X ∈ R) x = 3 √X f 1 (X )= 3 √X f 1 (x)= 3 √x (2) (3) f 1 f (x) = 3 f (x)= 3 √x3 = x; f f 1 (x) = f 1 (x) 3 = 3 √x 3 = x. f (x)=2x 1 1 R R+ f 1 R+ → R x ∈ R 2x = X (X ∈ R+ ), x = 2 X f 1 (x)= 2 x f 1 (f (x))= 2 2x = x 2 2= x (x ∈ R); f f 1 (x) =2 2 x = x (x ∈ R+ )
f (x)= x 2
+
1 1 1 x ∈ R x 2 = X (X ∈ R+ 0 ), ±√X f 1 (X )= ±√X x √x = √x y = x (a,b) f b = f (a) a = f 1 (b) (b,a) f 1 (a,b) (b,a) y = x y = f (x) y = f 1 (x)
= x f (x)= x 3
f (x)= 3 √x f (x)=2x f (x)= 2 x f (x)= x 3
f (x)= 3 √x f (x)=2x f (x)= 2 x f (x)= x 2 f (x)= √x f (x)= x 2 f (x)= x 2 1 1 R R+ 0 R+ 0 R f (x)= x 2 R
+ 0
+ 0 1 1
+ 0 R+ 0 f 1 R+ 0 → R+ 0 x ∈ R+ 0 x 2 = X (X ∈ R+ 0 ) (7)
X f 1 (x)= √x f 1 f (x) = √x2 = x (x ∈ R+ 0 ) (8) f f 1 (x) = √x 2 = x (x ∈ R+ 0 ) (9)
x ∈ R (8) (9) x ∈ R x ∈ R (√x)2 x< 0 f (x)= x 2 (10) f (x)= x 2 (x ∈ R+ 0 ) (11) f 1 (x)= √x (10) x 2 R f (x)= x 2 f (x)= x 2 (x ∈ R) (10) (11) (11) (10) R+ 0 y = x 2 f (x)= x 2 y = x y = ±√x y 2 = x (11) f 1 (x)= √x f (x)= x 2 (x 0) f 1 (x)= √x f (x)= x f (x)= x + α (α ∈ R) f (x)= ax + b cx a (a,b,c ∈ R) f 1 (x)= f (x) (12) f (x)= 1 2 3x 3 x
) 1 1
1 f (x)=3 4x
(x)= x +1 f (x)=1 x f (x)= 2 x
(x)= x +3 2x 1 f (x)= x x +1 f (x)= x 2 +1(x 0)
)= x 2 2x (x 1)
(x)= √x +1
(x)=1 √x +1 f (x)=2x +1 f (x)= 3x 1+3x f (x)= x 2 x
<pOq
(n ∈ N) <pOq <pOq <rO s <pOs <pOq + <rO s (n,< pOq )
n =0, 1, 2,..., < pOq
=0 (0,< pOq )
pOq
pOq < rO s < pOq + < rO s =(1,< pOs), (1,< pOq )+(2,< rO s)=(4,< pOs),
x,y ) (k ) x 2 + y 2 =1. (1) (1) (k ) (k )
ONM OPQ
<POQ = x = x 1 = |ON | |OM | = |OP | |OQ| ;
<POQ = y = y 1 = |MN | |OM | = |QP | |OQ| OPQ <pOq <pOq
x (90◦ x)+ x (90◦ x) x + (180◦ x)+ (90◦ + x) (π x) (π + x) Ä π 2 xä Ä π 2 + xä (2π x) ( x) ( x) (π + x) Ä π 2 + xä x + x Ä π 2 +
x
x
+
+
)
, π 2 x
M0 ,M1 ,...,M8 M0 = A M8 = B AB π 2
M0 M1 ,M1 M2 ,...,M7 M8 π 2 8 π 16 x kπ 16
(k =1, 2,..., 8)
16
16
p1 ,p2 ,...,p8 y pk x = kπ 16 (k =1, 2,..., 8) q1 ,q2 ,...,q8 M0 ,M1 ,...,M8 x qk y = kπ 16 (k =1, 2,..., 8) Sk pk qk Sk = kπ 16 , kπ 16 (k =1, 2,..., 8) s0 ,s1 ,...,s8 y = x
∈ [0,π ] ( x)= x
)=
∈ R
(x)= x i f (x)= x x ∈ R R [ 1, 1]
(x)= x
(x)= x
∈ R ( x)= x 1 x 1
(x)= x y =1 y = 1 x v x =0 ⇔ x = kπ (k ∈ Z f (x)= x kπ (k ∈ Z) x x> 0 0 <x<π, x< 0 π<x< 2π,
f (x)= x y
f (x)= x 1 v f (x)= x π 2 + kπ (k ∈ Z) x x x ∈ π 2 +2kπ
2 +2kπ x x ∈ π 2 +2kπ, 3π 2 +2kπ k ∈ Z
f (x)= x ((2k +1)π, (2k +2)π ) (2kπ, (2k +1)π ) k ∈ Z
f (x)= x x =2kπ 1 x =(2k +1)π k ∈ Z f (x)= x π 2 +2kπ, π 2 +2kπ π 2 +2kπ, 3π 2 +2kπ k ∈ Z f (x)= x f (x)= x x f (x)= x x ∈ 0,
M0 ,M1 ,...,M8
p1 ,p2 ,...,p8 t A N0 ,N1 ,...,N7 t CM0 ,CM1 ,...,CM7
n1 ,n2 ,...,n7 x N1 ,N2 ,...,N7
nk y = kπ 16 (k =1, 2,..., 7) Tk
pk nk Tk = kπ 16 , kπ 16 k =1, 2,..., 7 t1 ,t2 ,...,t7 y = x
Ä0, π 2 ä y = x
f (x)= x π 2 , π 2
f (x)= x π x = π 2 + kπ y = x
N7 n7
f (x)= x
f (x)= x π 2 + kπ π 2 + kπ k ∈ Z x = π 2 + kπ (k ∈ Z) R
f (x)= x π
f (x)= x
f (x)= x v f (x)= x kπ (k ∈ Z) x
f (x)= x x x ∈ kπ, π 2 + kπ x x ∈ π 2 + kπ,kπ k ∈ Z f (x)= x π 2 + kπ π 2 + kπ k ∈ Z
f (x)= x
kπ, π 2 + kπ π 2 + kπ,kπ k ∈ Z
f (x)= x
f (x)= x
f (x)= x f (x)= x Äx π 2 ä = x
f (x)= a (bx + c),f (x)= a (bx + c),f (x)= a (bx + c), a,b,c ∈ R
f (x)= x 2
f (x)= a(x α)2 + β
f (x)= a (bx + c) i f (x)= bx (b> 0)
Åx + 2π b ã = b Åx + 2π
ã = (bx +2π )= bx = f (x), (1)
π b (1) bx =0 ⇔ bx = kπ ⇔ x = kπ b (k ∈ Z).
f (x)= bx b> 0
f (x)= x b> 1 f (x)= x 0 <b< 1
y = x y = 3x
y = x y = x 2
f (x)= (bx + c)(b> 0,c ∈ R)
(bx + c)= b x + c b , (2) (1) |c| b c> 0
0 y = 3x y = (3x 1)
y = x 2 y = Ä x 2 +1ä
f (x)= a (bx + c)(a> 0,b> 0,c ∈ R)
= (3x 1) y =2 (3x 1) y = 1 2 (3x 1)
f (x)= a (bx + c)(a< 0,b> 0,c ∈ R) (3) x b< 0 a (bx + c)= a ( bx c), b> 0 f (x)= 2 (3x 1) f (x)=2 (3x 1) x
(1 3x)=
(1 3x)
(3x 1)
(x)= 1 2 (3x 1) x
f (x)= a (bx + c)(a,b,c ∈ R)
(x)=
x + π 4 ä = 1 √2 ( x + x) Äx + π 4 ä = 1 √2 ( x x) 2 Äx + π 4 ä Äx π 4 ä = 2 x 2 x 2 Ä π 4 + xä Ä π 4 xä = 2 x 2 x (x + y ) x y = x + y (x y ) x y = y x (y z ) y z + (z x) z x + (x y ) x y =0 x = 4 5 y = 3 5 0 <x<
x + y =
4 x Äx + π 4 ä =3 x + y = π 4 (1+ x)(1+ y )=2 x + y = π 2 (x y ) x y =
(1)
(3)
(5)
(6)
(1) (2) (5) (6)
(10)
Ä π 6 xäÄ π 6 + xä =8 Ä π 12 + x 2 ä Ä π 12 x 2 ä 2 Ä π 12 + x 2 ä Ä π 12 x 2 ä
=4 Ä π 6 + xä Ä π 6 xä .
(x)= x (0,π ) (π, 2π )
)
x
= 1 2 x π 4 = π 6 +2kπ,x π 4 =(2k +1)π
4
6 (k ∈ Z), (18) ]x = 5π 12 +2kπ,x = π 12 +(2k +1)π (k ∈ Z). (2x 5) (x +1)=0 4x = 5 3 4x x = 2x; 2x =2 x 5x + 3x =0 9x + 7x =0 2x 2 x =0 x + 2x +1=0 x 2 + x =1 1 x = x 2 x(1+2 x) 1=0 x 2x + 2x =0 2 x x x + x x =0 2 2 x 2 x +4 x 3=0 24 2 x 13 x 22=0 4 2 x 2(√2 1) x √2=0 x +3 x 4=0 2 x 4 x x +3 2 x =0 2 x + 2x 2 2 x = 1 2 2x + x = x + 2x 3x + x =0 2 2 2x 2 2 x =1 3x = 2 x 2x =3 3x x + 2x 3x =0 Ä π 4 xä + x =1
3x + 2x + x = 1 x + (x +2)= (x +1) x + 1= (x +1) x x =1 x + √3 x + √2=0 4x + √3 4x = √2 √3 x x √2=0 2 x +5 x =4 x +16 x =20 5( x + x)2 13( x + x)+8=0 x 2x = 3x (1+ 4x) 4x = 2 2x (x +1) 2(x +1)= 3(x +1) 4(x +1)
2x x = 4 3 2x 1+ x 1 x =1+ 2x 6x 2x = 5x 3x 2x
a = 1 (2) x x = 1 x = π 2 +2kπ (k ∈ Z) (3)
1 <a< 1 x (2) y = x y = a x (3) y = a y = x y = a x = a
2 , π 2 a α = a,β = π a,γ =2π + a. (4) (2) α<x<β (3)
β<x<γ x (2)
α +2kπ<x<β +2kπ (k ∈ Z), x (3)
β +2kπ<x<γ +2kπ (k ∈ Z) (4) (2)
2kπ + a<x< (2k +1)π a (k ∈ Z), (3)
(2k +1)π a<x< (2k +2)π + a (k ∈ Z).
1 <a< 1 y = x y = a x = a a [0,π ] α = a x>a
α<x<α, a<x< a, x<a
α<x< 2π α, a<x< 2π a.
1 <a< 1 x>a 2kπ a<x< 2kπ + a (k ∈ Z), x<a 2kπ + a<x< (2k +2)π a (k ∈ Z). a ≥ 1 a< 1 a y = x y = a
= a a π 2 , π 2 α = a x>a a<x< π 2 x<a π 2 <x< a x>a x kπ + a<x< π 2 + kπ (k ∈ Z),
(8)
(9) (8) (9) (7)
(10)
(11) (10) (11) (7)
(4) (7)
(12)
(7) (1)
(13)
(14)
(15)
(16)
(17)
i ax 2 + bx + c =0(a,b,c ∈ R,a =0)
x R x 2 5x +6=0 (2) x x =2 x x =3 x R x =2 x =3 x R x 2 5x +6 < 0 (3) x (2, 3) x (−∞, 2] ∪ [3, +∞) x1 + x2 =3 2x1 x2 =0 x1 x2 R x1 + x2 =3 ∧ 2x1 x2 =0 (4) x =(x1 x2 R2 (4) x (1, 2) x1 ,x2 )=(1, 2) x1 =1 x2 =2 x1 ,x2 ) =(1, 2) x1 =1 x2 =2 F (x) x S F (x) x x S F (x) x F (x) S F (x) F (x) x x x R(F ) F (x)
(x)
x ∈ R(F ), (5)
(x) S
(F )= S
(x) S (1) (4) (1) (4) {2} {(1, 2)} (2) {2, 3} (3) (2, 3) {x|2 <x< 3} x 2 +1=0 R x 2 +1 > 0 (x +1)2 = x 2 +2x +1 R (5)
(x) ⇔ x ∈ R(F ). (6) x 2 5x +6=0 ⇔ x ∈{2, 3} x 2 5x +6=0 ⇔ x ∈{2, 3} F (x) x (6) (6) F (x) R(F )
R x =1
x =2 ∨ x =3 (8) x 1 (9) [1, +∞) x =1 ⇔ x ∈{1} x =2 ∨ x =3 ⇔ x ∈{2, 3} x 1 ⇔ x ∈ [1, +∞) (7) (8) (9) (1) 7x 11=3 7x =14 x =2 x = 2 (1) x =2 x =2 (1) (1) {2} (1) x =2 (1) x ∈{2} (1) x =2
(x 2)(x 3)=0 x 2=0 ∨ x 3=0 x =2 ∨ x =3
(2) x =2 x =3 x =2 ∧ x =3 x =2 x =3
(2) x =2 ∧ x =3
(2)
(2) {2, 3} (2) x =2 ∨ x =3 (2) x ∈{2, 3} (2) x =2 x =3 (2) x =2 ∧ x =3 (2) 2∧3
x ∈{2, 3}⇔ x =2 x =3 {2, 3} x =2 x =3 x = ±2 x =2 x = 2 ±2 2 x =0 (10) {kπ | k ∈ Z} (10) x = kπ, k Z (11) (2) x = t t {2, 3} (2) x =3t 1 t 1, 4 3
(11) (10) (∀ k ∈ Z)(x = kπ ) (12) (10)
∃ k ∈ Z)(x = kπ ) (13) (12) x R (13) (10)
(G)
x S F (x) G(x) (15) x S
(F )= R(G) ⊆ S. (16) x 1 x 2 =0 x 1=0
\{2}
x =0 x 1=0
+ (x 1)(x 2) (x 2)(x 3) =0 x 1=0
\{2, 3} x 1 x 2 =0, 2 x =0, (x 1)(x 2) (x 2)(x 3) =0,x 1=0 (17) {1} (17)
(18)
\{1}
(19)
\{1, 2}
S S (x 1)(x 2)=0
+ \{2, 3}
(x 1)(x 2)=0
1)(x 2)
(26)
(x)=0 ⇒ g (x) =0. (27) (24) f (x)=0 ∧ g (x) =0 ⇔ f (x)=0, (28) (26) (28)
)
(x)=0 (29)
(x)= x 1
(x)
(x) g (x) =0 ⇔
x)= x 2 (27) (29) (21) (25) x ∈ S x R
(x) ∧ x ∈ S S (29) (27) f (x)= g (x)= x 2 (29) x 2 x 2 =0 ⇔ x 2=0, x =2 S F (x)
)
(30)
)
a n b
(0, +∞) (−∞, 0) (0, 0) (−∞, 0) (0, +∞) (0, 0) (0, +∞) (−∞, 0) (0, 3) (0, +∞) (−∞, 0) (0, 2) (0, +∞) (−∞, 0) (0, 1) (1, +∞) (−∞, 1) (1, 0) ( 3, +∞) (−∞, 3) ( 3, 1) (−∞, 2) (2, +∞) (2, 2) (−∞, 1) ( 1, +∞) ( 1, 3)
a
s
) (2, 1) (2, 1) ( 2, 1) ( 2, 1) (9, 7) (9, 7) ( 9, 7) ( 9, 7) (5, 1) ( 5, 1) (√13, √13)( √13, √13) (3, 1) ( 3, 1) (√2, 2√2) ( √2, 2√2) ( 6, 6) 6 5 , 3 5 (9, 9) 9 5 , 9 10 (5, 4) ( 5, 4) (4i, 5i) ( 4i, 5i) (1, 2) (2, 1) (1, 3) ( 3, 1) (2, 0) (0, 2) Å 3+ i√11 2 , 3 i√11 2 ã Å 3 i√11 2 , 3+ i√11 2 ã (1, 3) 1 5 , 3 5 (6, 1) (2, 3) 4, 3 2 (3, 2) ( 6, 3) Å 1+ i√143 2 , 1+3i√143 2 ã Å 1 i√143 2 , 1 3i√143 2 ã (0, 0) (1, 2) 15 22 , 9 22 (0, 0) Å 2+2i√2 3 , 2+ i√2 3 ã Å 2 2i√2 3 , 2 i√2 3 ã (0, 0) Å ab + b√ab a b , ab + a√ab b a ã Å ab b√ab a b , ab a√ab b a ã a 2 b2 4ab =0 Å ab + √a2 b2 4ab 2b , ab + √a2 b2 4ab 2a ã Å ab √a2 b2 4ab 2b , ab √a2 b2 4ab 2a ã a 2 b2 < 4ab =0 Å ab + i√4ab a2 b2 2b , ab + i√4ab a2 b2 2a ã Å ab i√4ab a2 b2 2b , ab i√4ab a2 b2 2a ã a =0 b =0 a =0 b =0 a = b =0 t, 1 t t =0 a 2 + b2 =0 Å ab2 a2 + b2 , a 2 b a2 + b2 ã a = b =0 (a +1,a 1) ( a 1, 1 a) 12 16 h2 4P h2 =4P a 2 P a2 + P ± a2 P 3= 2 8 1= 2 1 2 2= 3 1 9 1 3
f
=
y
x n (n ∈ N)
y =
A (ax + b) y = A (ax + b)
y = x n (n ∈ N)
=
y
A (ax + b) y = A (ax + b)
Издавач Завод за уџбенике Београд, Обилићев венац 5 www.zavod.co.rs
Ликовни уредник Тијана
Лектор
Графички уредник Александар Радовановић
Дизајн и прелом Жељко Хрчек
Корице Александар Радовановић
Коректор Добрило Тошић
Обим: 19,0 штампарских табака
Формат: 16,5×23,5 cm
Тираж: 00 примерака
Штампање завршено 20 . године. Штамп
