Question

1.Phân tích đa thức thành nhân tử

a) x² - y² + 4y - 4

b) (x-2)(x-1)(x+1)(x+2)-10

Answer 1

\(a,x^2-y^2+4y-4\)

\(=x^2-\left(y^2-4y+4\right)\)

\(=x^2-\left(y-2\right)^2\)

\(=\left(x-y+2\right)\left(x+y-2\right)\)

b, \(\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)-10\)

\(=\left[\left(x-2\right)\left(x+2\right)\right]\left[\left(x-1\right)\left(x+1\right)\right]-10\)

\(=\left(x^2-4\right)\left(x^1-1\right)-10\)

\(=x^4-x^2-4x^2+4-10\)

\(=x^4-5x^2-6\)

\(=x^4-6x^2+x^2-6\)

\(=x^2\left(x^2-6\right)+\left(x^2-6\right)\)

\(=\left(x^2-6\right)\left(x^1+1\right)\)

Answer 2

a.

\(x^2-y^2+4y-4\\ =x^2-\left(y^2-4y+4\right)\\ =x^2-\left(y-2\right)^2\\ =\left(x-y+2\right)\left(x+y-2\right)\)

Answer 3