Question

Phân tích thành nhân tử :

1) ( x² + 2x - 2 ) ( x² + 2x + 3 ) - 6 .

2) ( x² - 4x + 6 ) ( x² - 4x + 8 ) - 8 .

Answer 1

1) Đặt \(x^2\)+2x=t
Ta có \(\left(x^2+2x-2\right)\left(x^2+2x+3\right)-6\)=(t-2)(t+3)-6=\(t^2+t-6-6\)\(=t^2+t-12\)\(\left(t-3\right)\left(t+4\right)\)\(=\left(x^2+2x-3\right)\left(x^2+2x+4\right)=\left(x+3\right)\left(x-1\right)\left(x^2+2x+4\right)\)

2) Đặt \(x^2-4x+7=t\)
Ta có : (\(\left(x^2-4x+6\right)\left(x^2-4x+8\right)\)\(-8\)\(=\left(t-1\right)\left(t+1\right)-8=t^2-1-8=t^2-9=\left(t-3\right)\left(t+3\right)\)\(=\left(x^2-4x+4\right)\left(x^2-4x+10\right)\)\(=\left(x-2\right)^2\left(x^2-4x+10\right)\)