Question

Bài 2 phân tích đa thức thành nhân tử

a,x^2(x-1)+16(1-x)

b,5x^2-5y^2-10x+10y

c,x^2+4x-y^2+4

Answer 1

a) Ta có: \(x^2\left(x-1\right)+16\left(1-x\right)\)

\(=x^2\left(x-1\right)-16\left(x-1\right)\)

\(=\left(x-1\right)\left(x^2-16\right)\)

\(=\left(x-1\right)\left(x-4\right)\left(x+4\right)\)

b) Ta có: \(5x^2-5y^2-10x+10y\)

\(=5\left(x^2-y^2\right)-10\left(x-y\right)\)

\(=5\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\cdot2\)

\(=5\left(x-y\right)\left(x+y-2\right)\)

c) Ta có: \(x^2+4x-y^2+4\)

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

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

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