2. y = ax + b; a = 3; y = 3x + b; –4 = 3 ⋅ 2 + b; b = –10; y = 3x – 10. ПС-3 ⎛
1. ⎜ ⎜⎜ ⎝
(
b + c2
)
b − c2
(
)
b − c2 ⎞ ⎛ b ⎟:⎜ ⎟⎟ ⎜⎜ ⎠ ⎝
c2 b
( b + c ) − c ( b − c ) ⎞⎟ = b − c . ( b − c )( b + c ) ⎟⎟⎠ c b 2
2
2
2
2
4
2
2. 8(2 + 3y) + 3y(2 – 3y) = –8; 16 + 24y + 6y – 9y2 = –8; 9y2 – 30y – 24 = 0; 2⎞ 2 ⎛ ⎜ y + ⎟ (y – 4) = 0, т.к. 2 + 3y = 0 при y = − , то ответ: 4. 3⎠ 3 ⎝
ПС-4 ⎛
1 ⎞⎛
1⎞
⎛ 1 1⎞
1. 8x2 – 2x – 1 < 0; ⎜ x + ⎟⎜ x − ⎟ < 0 ; x ∈ ⎜ − ; ⎟ ; 8x2 – 2x – 1 ≥ 0; 4 ⎠⎝ 2⎠ ⎝ ⎝ 4 2⎠
(
)
x ∈ −∞; − 1 ⎤ ∪ ⎡ 1 ; +∞ . 4⎦ ⎣ 2 2
2. 9x – 10x + 1 =(9x – 1)(x – 1). ⎛ ⎝
1 ⎞⎛
1⎞
⎠⎝
⎠
x
1
= 20 x 2 + x − 1 = 0 . 3. ⎜ x + ⎟⎜ x − ⎟ = x 2 + − 4 5 20 20
ПС-5
1. a4=a1+3b; a13 = a1+12b; a13–a4 =9b=–13; b = 2. q =
b
2
b
1
=−
−13 37 , тогда a1=a4–3b = . 9 3
b 17 5 19 95 ;S= 1 =− ⋅ =− . 1− q 17 2 612 19
3. 0,4(428571) = 0,4 + S. S — геометрическая прогрессия с b1 = 0,0428571; q =
b 3 1 31 ; S= 1 = , тогда 0,4(428571) = . 1000000 70 1 − q 70
ПС-6
1. а)
2 − 2sin 2 α 1 2 − 2sin 2 α +1 = + tgα ctgα = 2 1 − cos 2α 2sin α sin 2 α
при α =
3π ; 8
1 = 4−2 2 ; 2 sin α − sin x ⋅ sin x (−ctgx) б) = ctg 2 x . sin x ⋅ sin x ⋅ tgx cos α cos x cos α = = =1 ; ⎛π α⎞ ⎛π α⎞ ⎛π ⎞ cos α 2cos ⎜ − ⎟ sin ⎜ − ⎟ sin ⎜ − α ⎟ ⎝4 2⎠ ⎝4 2⎠ ⎝2 ⎠ sin α + tgα tgα = + sinα ctgα = 1 + cosα. б) tgα tgα
2. а)
119