Bài 1 (6 điểm): Phân tích các đa thức sau thành nhân tử:
a) (a + b)2 – m2 + a + b – m
b) x3 + 6x2 + 12x – 8
c) x2 – 7xy + 10y2
d) x4 + 2x3 - 4x – 4
Bài 2 (4 điểm): Tìm x, biết:
a) (3x – 5)2 – (x +1 )2 = 0
b) (5x – 4)2 – 49x2 = 0
ĐÁP ÁN
Bài 1
a) (a + b)2 – m2 + a + b – m = (a + b + m)(a + b – m) + (a + b – m)
= (a + b – m)(a + b + m + 1)
b) x3 + 6x2 + 12x – 8 = (x – 2)3
Cách khác: x3 + 6x2 + 12x – 8 = (x3 – 8) - 6x2 + 12x
= (x – 2)(x2 + 2x + 2) + 6x(x – 2) = (x – 2)( x2 + 2x + 2 – 6x) = (x – 2)3
c) x2 – 7xy + 10y2 = x2 – 2xy – 5xy + 10y2 = x(x – 2y) – 5y(x – 2y)
= (x – 2y)(x – 5y)
d) x4 + 2x3 - 4x – 4 = (x4 – 4) + (2x3 – 4x) = (x2 – 2)(x2 + 2) + 2x(x2 – 2)
= (x2 – 2)(x2 + 2 + 2x) = (x - √2)( x + √2)( x2 + 2 + 2x)
Bài 2
a) (3x – 5)2 – (x +1 )2 = (3x – 5 – x – 1)(3x – 5 + x + 1)
= (2x – 6)(4x – 4) = 8(x – 1)(x – 3)
Vậy (x – 1)(x – 3) = 0 ⇒ x - 1 = 0 hoặc x - 3 = 0
⇒ x = 1hoặc x = 3
b)(5x – 4)2 – 49x2 = (5x – 4)2 – (7x)2 = (5x – 4 – 7x)(5x – 4 + 7x)
= (12x – 4)(-2x – 4) = -8(3x – 1)(x + 2)
Vậy (3x – 1)(x + 2) = 0 ⇒ 3x - 1 = 0 hoặc x + 2 = 0
⇒ x = 1/3 hoặc x = -2