MATEMATIKA
za II razred sredwe {kole
Редактор
f (x)= xn
f (x)= x 2
f (x)= x 3
f (x)= xn (x ∈ N) xn = a
f (x)= ax 2
f (x)= a(x α)2 f (x)= ax 2 + β
)=
⇒ a<b2 ,
(x)= x n
0
2 f (x)= x 2 x> 0
)=
)=
(x,x 3 ) Oxy
f (x)= x 3 y = x 3 ( x)3 = x 3
x> 0 x x< 0 x
x> 0 ⇒ x 3 > 0; x< 0 ⇒ x 3 < 0; a b
a<b a 3 <b3 a b a<b a 3 <b3 a b
a<b ⇒ a 3 <b3
f (x)= x 3 x ∈ R
f (x)= xn (x ∈ N)
f (x)= xn , n n =2 n =3
f (x)= x 4 f (x)= x 6 ,...
f (x)= x 2k (k ∈ N)
f (x)= x 2 f (x)= x 5 f (x)= x 7 ,...
f (x)= x 2k +1 (k ∈ N)
f (x)= x 3 y = xn
f (x)= x n
f (x)= xn (n ∈ N) n x 0
f (x)= xn x> 0 ⇒ xn > 0 0 a<b ⇒ an <bn
a = b ⇔ an = bn (a> 0,b> 0). a = b an = bn
n = bn a b
n
n = a · a ··· a. an n n 1 2 am an = am+n , (am )n = amn , (ab)n = an bn am an = am n (m>n; a =0) m = n
m =0 a0 an = a 0 n a 0 =1 1 an = a n an a =0 n n 0 a =0 0n = 0(n 1) (n 0) a n an an n m>n m = n m =0 a 0 a n m n m n m n a =0 b =0
4 a 3 = a4 a3 = a, a 4 a 3 = a 4 3 ,
5 a 2 = a2
5 = 1
3 = a 3 , a 5 a 2 = a 5+2 , a 5 a 3 = 1 a5 1 a3 = 1 a8 = a 8 , a 5 a 3 = a 5+( 3) , (a 3 ) 2 = 1 (a3 )2 = 1 a6 = a 6 , (a 3 ) 2 = a 3·( 2) , (a 4 )3 = 1 (a4 )3 = 1 a12 = a 12 , (a 4 )3 = a( 4) 3 ,
a> 0 √a √a
a √a ±√a a =0 x 2 =0 a< 0 x ∈ R x 2 0 x x 2 = a< 0 n a> 0 n √a n √a a =0 a< 0 n f (x)= xn xn = a y = xn y = a
a> 0 y = xn y = a n √a n √a xn = a a =0 y = xn y = a y =0 xn =0
a< 0 y = xn y = a xn = a x 4 =16 2 x 6 =20 6 √20 6 √20 x 8 =0 x 2 = 1 n =3 x 3 = a.
= B ⇔ An = B n , (A,B> 0; n ∈ N),
a m n = a p q .
x = m n m ∈ Z n ∈ N ax a m n a b m p n q a m n · a p q = a m n + p q ; a m n a p q = a m n p q ; a m n p q = a m n · p q ; (ab) m n = a m n b m n ;
b m n = a m n b m n . A = a m n · a p q ,B = a m n + p q ,
nq = a m n · a p q nq = a m n nq a p q nq nq ∈ N = amq · apn = amq +pn nq,pn ∈ Z ; B nq = a m n + p q nq = a mq +pn nq nq = amq +pn nq ∈ N
x 2 = a a x 2 0 ∈ R x 2 = a> 0 x x 2 = 1; R
i i2 = 1 i a b ,... ai = ia,a + i = i + a, (a + b)i = ai + bi, (a + bi)i = ai + bi2 = ai b i2 = 1 (a + bi)c = ac + bci,i3 = i2 · i =( 1) · i = i, i x x 2 0 i i2 = 1 ( i)2 = 1 ( i)2 =( 1)2 i2 =12 i2 = i2 = 1. i i 1 a a 2 = 1 i2 = 1 a 2 = i2 , a 2 i2 =0, ι i i j i
a,b,c,d ∈ R
a + bi (a,b ∈ R
a + bi = c + di ⇔ a = c ∧ b = d.
a = c ∧ b = d a + bi = c + di a = c a + bi = c + di a c = di bi, a c =(d b)i. i i(a c)=(d b)i2 = b d, a c =0 i = b d a c .
i / ∈ R
b = d a + bi = c + di i = a c d b , a = c b = d a + bi = c + di
a + bi c + di A + Bi (A,B ∈ R) C
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)= a + 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+0i
c d
a + bi c + di = a + bi c + di · c di c di = ac adi + bci bdi2 c2 cdi + cdi d2 i2 = (ac + bd)+(bc ad)i c2 + d2 i2 1 = ac + bd c2 + d2 + bc ad c2 + d2 i.
ac + bd c2 + d2 bc ad c2 + d2 ∈ R i2 = 1,i3 = i2 i =( 1)i = i,i4 =(i2 )2 =( 1)2 =1, i5 = i4 · i =1 · i = i,i6 = i4 · i2 =1 · ( 1)= 1, i7 = i4 · i2 · i =1 · ( 1) · i = i,i8 =(i4 )2 =12 =1, in n =1, 2, 3,... {in | n ∈ N} = {i, 1, i, 1}. (1+ i)2 (1 i)3 (1+ i)4 (1+ i)2 =1+2i + i2 =1+2i 1=2i, (1+ i)3 =1+3i +3i2 + i3 =1+3i 3 i = 2+2i, (1+ i)4 =((1+ i)2 )2 =(2i)2 = 4
(1+ i)(2 i) 2+3i 2+3i (2 i)(1+ i)
∈ C (iz )= z (iz )= z
+ z =0 ⇔ (iz )= (iz )
x 4 =1 x 4 1=(x 2 1)(x 2 +1))
z = a + bi
Oxy z =(a,b) 2+
3i 1+2i 2 i 1 2i (2, 3) ( 1, 2) (2, 1) (1, 2)
a a + 0i (a, 0) x bi 0+ bi (0,b) y i =0+1 · i (0, 1) i 0 1 · i (0, 1)
z = a + bi (a,b ∈ R) z = a bi
z = a + bi
z = a bi
z = z x
z = a + bi z = a bi
z z z = a + bi (a,b ∈ R)
z · z =(a + bi)(a bi)= a 2 + b2 .
2 x x +4 x(x +4) x(x +4)=60, x 2 +4x 60=0. x ax 2 + bx + c =0, a b c a =0 a =0 bx + c =0 2x 2 x 8=0 x 2 = 1 2x x 2 =5 x 2 = x x 2 +1=0 x 2 +2x 5=0 x 2 x =0
AB =0 ⇔ A =0 ∨ B =0. x =0 ax + b =0 x =0 x = b a x 2 +3x =0 x(x +3)=0, x =0 ∨ x +3=0, 3 3x 2 =0 7x 2 +3=0 7x 2 1=0 8x 2 +12=0 3x 2 4x =0 x 2 +2x =0 (x +1)2 =4 2 x = x 7 m ax 2 + c =0 3x 2 15x +5mx =18 x + m x 2 = 5x mx 2 x m 4 = 5 x 3x + m x 2 + 1 2 = m + 1 2 (a + bx)2 +(ax b)2 =2(a 2 x 2 + b2 ) x + m m x = n + x x n 2 x 3 6 x 5 + 3 x 4 =0 ax 2 + bx + c =0, (a =0). ax 2 + c =0,
x 2 +8x 16=0
2x 2 x 6=0
3x 2 = x +1 1 3 (x 2 + x)=1 x +3 x 3 + x 3 x +3 =3 x x a + x a x = 5 2
1 m + 1 m + x + 1 m +2x =0 1 x a + 1 x b = 1 x c a (1 a)x 2 4ax +4(1 a)=0 m x 2 +(m 1)x +36 (x + α)2 ax 2 + bx + c =0
f (x)= x +1 x 1 2 4 x 1 x +1 2
f (x)=0 f (x)=4
ax 2 + bx + c ax 2 + bx + c = a(x α)(x β ) b2 4ac> 0 b2 4ac =0 b2 4ac< 0
a (1 a)2 +4(2 a)=1 2a + a 2 +8 4a = a 2 6a +9=(a 3)2 0, a ∈ R α β αβ = (2 a), (2 a) > 0 2 a< 0 a> 2 α + β = a 1 > 2 1 > 0 a< 2 αβ = (2 a)= a 2 < 0 a =2 x 2 x =0 x =0 6x 2 x 1 x 2 x +1 2x 2 7ax +6a 2 x 2 5x +4 4x 2 5x +1 x 2 3ax +2a 2 ab b2 12a2 a 1 3a2 +5a 2 t2 9tu +14u2 t2 tu 2u2 3x2 11x +10 (5 x)3 8x3 a 4x 2 15x +4a 3 =0 x 2 (m +1)x + m =0 m m 4x 2 (3m +2)+(m 2 1)=0 a x 2 +(2a 1)x + a 2 +2=0 x 2 8x +12=0 a (a 2)x 2 4(a +2)x +4a =0
a b c a =0 f x ax 2 + bx + c x f (x)= ax 2 + bx + c x P (x) P (x)= x 2 s t
s(t)= 1 2 at2 + v0 t + s0 a v0 s0
f (x)= ax 2 + bx + c y = ax 2 + bx + c
f (x)= ax 2
f (x)= x 2 y = x 2 y = x 2
f (x)= x 2 x ∈ R f (x) 0 x x y = x 2 ( x)2 = x 2
y f f ( x)= f (x)
f (x)= x 2
x f (x)= x 2 x
f (x)= x 2
f (x)= x 2 x> 0 x< 0 f
(A,B ) x1 <x2 f (x1 ) <f (x2 ) x1 x2 x1 x2 x1 < x2 f (x1 ) >f (x2 )
f (x)= x 2 (0, +∞) (−∞, 0)
x1 x2 x1 <x2 x1 <x2 x1 x2 x 2 1 <x1 x2 x1 x2 <x 2 2 , x2 x 2 1 <x 2 2
f (x)= x 2 x1 <x2 ⇒ f (x1 ) <f (x2 ) x1 x2 x> 0 (0, +∞) x1 x2 x1 <x2 (< 0) x1 x2 x 2 1 >x12 x1 x2 >x 2 2 , x 2 1 >x 2 2 f (x)= x 2
(−∞, 0) (0, 0) f (x)= x 2 x =0 f (x) >f (0)=0 (0, 0) y = x 2
(0, 0) f (x)= x 2 y = x 2
x 2 f (x)= x 2 x1 x2 (x1 ,x 2 1 )
(x2 ,x 2 2 ) y = x 2
f (x)= x 2
f (x)= x 2
(x0 ,x 2 0 )
f (x)= x 2 x
f (x)= x 2 y = x 2
x =0
f (x)= x 2
f (x)= x 2 (x0 , x 2 0 )
x ∈ R x 2 0 x
f (x)= x 2 ( x)2 = x 2 y
f (x)= x 2 x< 0 (−∞, 0) x> 0 (0, +∞)
(0, 0)
f (x)= x 2 x =0 f (x) <f (0)=0 (0, 0) y = x 2
f (x)= x 2
f (x)= x 2 x ∈ R y = x 2 y = x 2 y = x 2
f (x)= ax 2 a =0 a> 0 f (x)= ax 2
f (x)= x 2 y = x 2 ax 2 >x2 a> 1 ax 2 <x2 0 <a< 1 y = ax 2 a a y = x 2 y =2x 2 y = 1 2 x 2
f (x)= ax 2 a< 0 f (x)= x 2 a |a| a |a|
(x)= a(x α)2
= x 2 y = 2x 2
(x)= ax 2 + β
(x)= a(x α)2 (α> 0).
1 =(x0 ,y0 )
= 1 2 x
= a(x α)2
0 = a(x0 α)2 M2 =(x0 α,y0 )
= ax 2
f (x)= ax 2 f (x)= a(x α)2 (a =0)
(α, 0)
a> 0 a< 0 (α, 0) y a> 0 x>α x<α a< 0 x>α x<α a> 0 (α, 0)
0 (α, 0)
)=
y = ax 2
(0, 0)
y = ax 2 y β β> 0 β β< 0
y = ax 2 x
y = ax 2 + β (0,β ) a> 0 y = ax 2 + β y = β (0,β ) x a< 0
a> 0 β> 0 y = ax 2 + p
y = β> 0 x a< 0 β< 0 y = β< 0 a> 0 β< 0 a< 0 β> 0
y = ax 2 + β x
β
y = ax 2 + β x
y = ax 2 + β ∧ y =0. ax 2 + β =0,
β y = ax 2 + β x
a < 0 β a > 0
a
a , 0
a , 0 y = ax 2 + β x f (x)= ax 2 + β f (x)= a(x α)2 y = ax 2 + β a> 0 β< 0
ax 2 + β> 0
f (x)= ax 2 + β
ax 2 + β< 0
β> 0
β a <x< β a
f (x)= ax 2 + β
y = ax 2 + β
f (x)= x 2 + β
a< 0
β a <x< β a x> β a x< β a
y = ax 2 + β (0,β )
f (x)= ax 2 + β
a> 0 (0,β )
a< 0 f (x)= ax 2
a> 0
a< 0
= 2x 2 +4
= 2x 2 (0, 4) 2x 2 +4=0 ⇔ x = ±√2 x ( √2, 0) (√2, 0) f (x)= 2x 2 +4 x √2 <x< √2 x x> √2 x< √2 y =3x 2 +2 (0, 2) y =2 x f (x)= 3x 2 +2
=0 (x0 ,y0 )
=2(x 3)2 +1
=2x 2 (3, 1)
b2 4ac< 0 a> 0
ax 2 + bx + c> 0 x ∈ R
b2 4ac< 0 a< 0
ax 2 + bx + c> 0 x ∈ R
+
+ √D 2a , ax 2 + bx + c> 0
+ √D
a <x< b √D 2a , ax 2 + bx + c< 0 x>
√D 2a x< b + √D 2a .
± √D 2a aβ> 0 b2 4ac< 0 aβ< 0 b2 4ac> 0 aβ =0 β =0 a =0 f (x)= a(x α)2 b2 4ac =0 ax 2 + bx + c = a(x α)2 > 0 a> 0 0 a< 0
b 2a ax 2 + bx + c =0 b2 4ac =0
+
(x)= x 2 +2x 3
± √4+12 2 3 x> 1 x< 3 3 <x< 1 y = x 2 +2x 3 x = 1 3 2 = 1 x> 1 x< 1 y = x 2 +2x 3 ( 1, 4) x 2 +2x 3=(x +1)2 4 y = x 2 x y f (x)= 2x 2 +4x 5 2x 2 +4x 5=0 x 2 2(< 0) y = 2x 2 +4x 5 (1, 3) x< 1 x> 1
2 +2x 3=0
2x 2 +4x 5= 2(x 1)2 3 y = 2x 2 x y f f (x) 2x 2 +1 1 2 x 2 2 (x 2)2 2 x + 1 2 2 +1 2x 2 3x 2 3x 2 +8x +3 p y = x 2 2x +13 a b c f (x)= ax 2 + bx + c 8 x =4 y = ax 2 + bx + c (6, 12) x (8, 0) a b c y = ax 2 + bx + c y (0, 15) ( 2, 7) a b c a b c f (x)= ax 2 + bx + c x = 1 2 f (0)=24 a 4x 2 4ax + a 2 2a +2 a b c x ax 2 + bx + c> 0 ax 2 + bx + c< 0; ax 2 + bx + c 0; ax 2 + bx + c 0. f (x)= ax 2 + bx + c a D = b2 4c
D< 0 a> 0 x ∈ R
x 2 +2x 3 > 0 x 2 +2x 3=0 3 y = x 2 +2x 3 x< 3 x> 1 (−∞, 3) ∪ (1, +∞)
2 + x +1 > 0
∈ R y = x 2 + x +1 x R
2 + x +1 < 0
ax + b 0 ax + b< 0 ax + b 0 ax> b x x> b a a> 0 x x< b a a< 0 x2 x 2 5 x > 0. x2 x 2 5 x x 2 x 2 > 0 5 x> 0 x 2 x 2 < 0 5 x< 0 x 2 x 2 > 0 ∧ 5 x> 0, x 2 x 2 < 0 ∧ 5 x< 0. f (x)= x 2 x 2 1 x> 2 x< 1 x< 5 x< 1 2 <x< 5 1 <x< 2 5 <x x x x< 1 2 <x< 5 (−∞, 1) ∪ (2, 5)
m f (x)=(m +1)x 2 2(3m 2)x +(m +1) x f (x)= ax 2 + bx + c x ∈ R
a> 0 ∧ b2 4 < 0 x ∈ R m +1 > 0 ∧ (3m 2)2 (m +1)2 < 0, m +1 > 0 ∧
+1 > 0
3 2 x 2 4x +3 > 0 x x 2 > 0 3x 2 +2x +1 > 0 3x 2 2x +5 < 0 x2 6x +18 x 4 > 0 x2 +5x +4 x2 5x 6 < 0 2 x 3 x 2 > x 2 x 1 14x x +1 9x 30 x 4 < 0 m x 2x 2 +(m 3)x +3 m> 0
(m 1)x 2 (m +1)x +(m +1) > 0
(m +1)x 2 3mx +4m< 0
(m 2)x 2 +2(2m 3)x +5m 6 < 0 m
(m 2)x 2 (3m 6)+6m =0
(x 1)(x 3)+ m(x 2)(x 4)=0
(m 2 1)x 2 +2(m 1)x + 1 2 =0
5m 2 3mx +1=0 m x
f (x)=(2m 5)x 2 +6(m 3)x +3(m 5)
f (x)= ax 2 + bx + c b 2a , 4ac b2 4a
a m n m = √2x n = √2(a x) P x P (x)= √2x√2(a x)=2ax 2x 2 P (x)=2ax 2x 2
(x,y )
ax + by + c =0
Ax2 + Bxy + Cy 2 + Dx + Ey + F =0 a b c A B C D E F a b A B C x y x y A = B = C =0
ax + by + c =0 ∧ Ax2 + Bxy + Cy 2 + Dx + Ey + F =0. (x,y )
ax + by + c =0 ∧ Ax2 + Bxy + Cy 2 + Dx + Ey + F =0, a b c A B D E F a b A B C a b b =0 ax + by + c =0 ⇔ y = a b x c b , ax + by + x =0 ∧ Ax2 + Bx a b x c b +C a b x c b 2 + Dx + E a b x c b + F =0.
x y +1=0 ∧ x 2 +3xy y 2 +2x 5y +64=0 x = y 7, x y 7 (y 7)2 +3(y 7)y y 2 +2(y 7) 5y +64=0, 3y 2 38y +99=0 11 3 y =9 x =2 y = 11 3 x = 10 3 (2, 9) 10 3 , 11 3
= C =0
+ by + c =0 ∧
2 + Dx + Ey + F =0
=0 x + y =2 ∧ y +2x 2 =3, y = x +2 y = 2x 2 +3 A =(1, 1) B = 1 2 , 5 2 (1, 1) 1 2 , 5 2
x + y =4 ∧ x 2 2y 2 xy +5x +15=0
xy =5 ∧ x 2y =3
x + y x y = 3 7 ∧ xy +2y = 14
3x 1 x +1 + 1 y 1 = y +2 (y 1)(x +1) ∧ 4y x =7
2x 2 3xy +4y 2 5x =43 ∧ 2x 3y =2 1
x +2y + y =2 ∧ y x +2y = 3 1
x y + y =1 ∧ x x y = 2 x 2 +2xy + y 2 x y =6 ∧ x 2y =3
2x + y 4=0 ∧ x 2 y 2 +4x 6y +2xy +7=0
x 2 +2y =10 ∧ x y = 1 2x 2 y = 2 ∧ 3x + y =1 x 2 y =8 ∧ x y =2 x y m m =10x + y 10x + y x + y =6+ 2 x + y ⇔ 10x + y =6(x + y )+2, 10x + y xy =5+ 2 xy ⇔ 10x + y =5xy +2. x + y =0 xy =0 4x 5y =2 ∧ 10x + y 5xy =2 (3, 2) 1 5 , 6 25 x y m =10 3+2=32
4s a
2x 2y 4x +4y =4s x 2 + y 2 = a 2 x + y>a s>a
x + y = s ∧ x 2 + y 2 = a 2 y s x 2x 2 2sx + s 2 a 2 =0
D =4s 2 8(s 2 a 2 )=4(2a 2 s 2 ) 0, d =2a 2 s 2 0 x = s ± √d 2 , y (= s x)
a<s a√2
2x 2y 2x = s + √d 2y = s √d
2y = s + √d 2x = s √d s = a√2 d =0 x = y 2
(x)= ax Q
=2 f (x)=2x (x, 2x )
(x)=2x y =2x
f (x)=2x Q f : Q → R+ x> 0 f (x)=2x > 1 x< 0 0 < 2x < 1 x =0 f (0)=20 =1 x = p q p q 2 > 1 2p > 1 q √2p > 1
√2p 1 q 2p 1 2p > 1 x< 0 x> 0 2 x > 1 2x 2x 2 x > 2x
2x < 1 y =2x
f (x)=2x x1 x2 x1 <x2 r (= x2 x1 ) x1 + r = x2 2x2 =2x1 +r =2x1 2r > 2x1 , 2r > 1 x1 <x2 ⇒ 2x1 < 2x2 , f (x)=2x
g (x)=3x
x> 0 g (x) >f (x) x< 0 g (x) <f (x)
g (0)= f (0)=1 x> 0 2x < 3x x< 0 2x > 3x
a> 1 y = ax
y =2x y =3x f (x)= ax
f (x)=2x
f (x)= ax x ∈ Q ax > 0
x ∈ Q x> 0 f (x)= ax > 1 x< 0 f (x)= ax < 1 f (0)= a 0 =1
f (x)= ax x1 ,x2 ∈ Q x1 <x2 ⇒ ax1 <ax2 . f f (x)= ax 0 <a< 1 a = 1 2 x 3 2 1 0 1 2 3 1 2 x 8 4 2 1 1 2 1 4 1 8
x, 1 2 x a = 1 3 0 <a< 1 f (x)= ax f (x)= ax > 0 x ∈ Q x> 0 f (x)= ax < 1 x< 0 f (x)= ax > 1 f (0)= a 0 =1 x1 ,x2 ∈ Q x1 <x2 ⇒ ax1 >ax2 f (x)= ax a> 1 0 <a< 1 a =1 f (x)=1x =1 x ∈ R x f g h k f (x)=10x g (x)= 10 x h(x)= 10x k (x)= 10 x f
f (x)=2 · 3x f (x)=2 · 3x 1 f (x)= 3 2x +1 f (x)= 3 2 x +3 0 <a<b< 1 x> 0 ax <bx x< 0 ax >bx x f
f (x)= 2x ,x 0 2 x ,x< 0. f (x)=2x x ∈ Q (x, 2x ) (x, 2x ) x ∈ Q 2x x ∈ Q x √2 √3 π
21,4142 < 2√2 < 21,4143 . f (x)=2x x ∈ Q x ∈ R x √2 y y =2x A x A y
2√2 A = (√2, 2√2 )
2x x (x, 2x ) x y =2x (x, 2x ) x f (x)=2x f (x)= ax (a> 0)
f (x)= ax a> 1 x ∈ R ax > 0 x ∈ R x> 0 ⇒ ax > 1 x< 0 ⇒ 0 <ax < 1 x1 <x2 ⇒ ax1 <ax2 ax+y = ax ay (ax )y = axy ax bx =(ab)x a b x y x1 = x2 ⇔ ax1 = ax2 . x1 = x2
x1 = ax2 x1 = x2 x1 <x2 x1 >x2 x1 <x2 ax1 <ax2 x1 >x2
a> 0 a = x ax = b
b> 0
b 0 a> 1 0 <a< 1
f (x)= ax b> 0 x ax = b a> 1
b1 b2 x1 x2
ax1 = b1 xx2 = b2 0 <a< 1
b 0 x ax = b ax > 0
ax = b 0 a b b a
(x, 2 x)
f (x)= 2 x y = 2 x a> y = a x y = 2 x y = 2 x y = 3 x
1 2 x 3 2
3 f (x)= 1 2 y = 1 2 x y = a x 0 <a< 1 y = 1 2 x y = 1 3 x
x2
1 <x2
(x> 1) a x x (x< 1) a x f (x)= a x 0 <a< 1
f (x)= 10 x f (x)= 1 4 x
f (x)= 2 (2 x)
f (x)=3 3 (x +1)
f (x)= 1 2 |x|
f (x)= 1 2 x
f (x)= 2 x + | 2 x|
f (x)=1+ 1 3 |x 1|
f (x)= 2 x2 | 2 x|
y = a x ⇔ ay = x. y = xn ⇔ x = n √y (x> 0,y> 0). A B f A B x1 ,x2 ∈ A
f (x1 ) f (x2 ) x1 = x2 ⇒ f (x1 ) = f (x2 )(x1 ,x2 ∈ A). y ∈ B x ∈ A y x f y = f (x)
f : A → B x1 = x2 ⇒ f (x1 )= f (x2 )(x1 ,x2 ∈ A), x1 = x2 ⇔ f (x1 )= f (x2 )(x1 ,x2 ∈ A). (p ⇒ q ) ⇔ (¬q ⇒¬p) p x1 = x2 ,q f (x1 ) = f (x2 ),
¬p x1 = x2 , ¬q f (x1 )= f (x2 ), f (x1 )= f (x2 ) ⇒ x1 = x2 f 1 1 x1 = x2 ⇔ ax1 = ax2 , f (x)= ax 1 1 f (x)= x 2 x1 = x2 ⇒ f (x1 )= f (x2 ), 2 f (x)= x 2 1 1 B x
y = f (x) B f
A
f f (x)= 2 x R+ R y ∈ R x y = 2 x f (x)= x 3 R R f (x)=2x R R 1 ∈ R x ∈ R 2x = 1 R R+ f (x)= x 2 R R R R+ 0 f 1 1 A B y ∈ B x ∈ A y = f (x)
: B → A y ∈ B x ∈ A
= f (x) ⇔ x = g (y )(x ∈ A,y ∈ B ). f A
g f f 1 y = f (x) ⇔ x = f 1 (y )(x ∈ A,y ∈ B )
x ∈ A f y = f (x) ∈ B f 1 x f 1 (y )= x f 1 (y )=
f 1 (f (x))= x y ∈ B x ∈ A f 1 (y )= x
f (x)= y f (f 1 (y ))= y f 1 1 A B
f 1 : B → A
f 1 (f (x))= x (x ∈ A)
f f 1 (f (y )) = y (y ∈ B ).
f 1 f f f 1 f f 1
f (x)=2x +3 1 1 R R f 1 f 1 (2x +3)= x. 2x +3= X, x = X 3 2 . f 1 (X )= X 3 2 , f 1 (x)= x 3 2 , f (x)=2x +3
f (x)= ax a> 0 a =1 1 1 R R+ f 1 R+ → R
f 1 (ax )= x. ax = X x = a X f 1 (X )= a X
f 1 (x)= a x.
f (x)= ax f (x)= a x f f 1 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) y = x
f f 1
f (x)=2x f (x)= 2 x
f (x)= x 2 1 1 R R+ 1 1 R+ R+
f (x)= √x
f (x)=3+2 f (x)= x 2 (x 0) f (x)= 1 2 x
f (x)= 1 2x f (x)= 3 (x 1) f (x)= x 3 f (x)= x 4
2 A2 SA3 An SA1 A1 A2 ,...,An n n
SAi (i =1, 2,...,n)
SAi (i =1, 2,...,n)
◦ n n 2 n · 180◦ m m · n 2 n · 180◦
◦ m · n 2 n · 180◦ < 360◦ . mn 2m< 2n 1 n + 1 m > 1 2
n = m =3 n =3 m =4 n =3 m =5 n =4 m =3 n =5 m =3 np =2i mt =2i t + p = i +2
n =3 m =3 ⇒ t =4 i =6 p =4
n =3 m =4 ⇒ t =6 i =12 p =8
n =3 m =5 ⇒ t =12 i =30 p =20
n =4 m =3 ⇒ t =8 i =12 p =6
n =5 m =3 ⇒ t =20 i =30 p =12
1 B1 C1
S
B B1 M
a1 (a>a1 )
P = B + B1 + M =64+4+100=168.
◦ s
◦
n 3
e
en =1, e = 1 n V (E )= V (Q) n3 = V (Q)e 3 = e 3 ,
V (Q)=1 Q e a b c O e e a p b q c r e a p e pe a< (p +1)e,qe b< (q +1)e,re c< (r +1)e.
pqr
(p +1)(q +1)(r +1) pqre 2 V (F ) < (p +1)(q +1)(r +1)e 3 . pqre 3 abc< (p +1)(q +1)(r +1)e 3 , (p +1)(q +1)(r +1)e 3 < abc pqre 3 pqre 3 (p +1)(q +1)(r +1)e 3 <V (F ) abc< (p +1)(q +1)(r +1)e 3 pqre 3 , |V (F ) abc| < (p +1)(q +1)(r +1)e 3 pqre 3 , |V (F ) abc| < (pq + pr + qr + p + q + r +1)e 3 ,
V (F )= BH.
V (F )= abc a b c a b
B = ab H c
V (F )= abc
F1
V (F )= V (F1 )= BH
H
ABC )H
Π( ABC )H V (F )= BH B
1 ,F2 ,... Fn 2
B1 ,B2 ...,Bn 2 n
2
(1, 4) (4, 1)
k
F1
F = SABC
F F1 ,F2 ,...,Fn
F2 ,...,Fn
Fk (k =2,...,n)
k
k
k
1 P1 n =4
P2 P3 P4 F2 F3 F4
P1 P2 P3 P4
F1 F2 F3 F4
2 ,...,Pn
1 ,P2
n
V (P2 )+ V (P3 )+ ··· + V (Pn ) <V (F ) <V (P1 )+ V (P2 )+ ··· + V (Pn ).
V (F )= 1 3 BH, B H Fn Fn n> 3 n 2 F1 ,F2 ,...,Fn 2
)=
(F )= 1 3 BH + 1 3 (B B )x = 1 3 BH + 1 3 √BB + B H,
n n
n Fn
n H Gn Ωn H n =4 Fn Gn F4 G4
n F F Gn
V (Fn ) <V (F ) <V (Gn ), n ∈ N V (T ) Bn ωn Fn Dn Ωn Gn Bn H<V (F ) <Dn H n ∈ N
2 n πr 2 H
(F )= πr 2 H.
M =2πrH.
B = πr 2
P =2B + M =2πr 2 +2πrH =2πr (r + H ). Q 2R H Q =2RH M =2RπH M = Qπ
r F n ωn n Ωn Fn Gn sn Ωn F n =4 n ∈ N
V (Fn ) <V (F ) <V (Gn ),
V (F ) Bn ωn Fn Dn Gn
V (Fn )= 1 3 Bn H,V (Gn )= 1 3 Dn H, 1 3 Bn H<V (F ) < 1 3 Dn H.
(
(
= πa
V = 4 3 πR 3 , P =4πR2 . F K R h
V (F )= 1 3 πh2 (3R h), Π(K )=2πRh,
S h
h1 h2 (h1 h2 = h)
Π(S )=2πRh1 2πRh2 =2πRh.
V (I )= V1 + V2 , V1 h V2
V (I )
V1 = 1 3 πh2 (3R h),V2 = 1 3 πr 2 (R h), OAS r 2 = R2 (R h)2 , r 2 =2Rh h2 ,
V2 = 1 3 π (2Rh h2 )(R h)= 1 3 πh(2R h)(R h).
V = V1 + V2 = 1 3 πh(3Rh h2 +2R2 2Rh Rh + h2 )= 2 3 πR2 h. h>R
V = V1 V2
H 2 = s 2 (R r )2 , H 2 =(R + r )2 (R r )2 , H 2 =4Rr, H =2√Rr. p H 2 = √Rr
=4ρ 2 π =4Rrπ. M = πs(R + r )= π (R + r )2
: M =4Rr :(R + r )2
y = xn (n ∈ N)
