Question

Phân tích đa thức thành nhân tử: \(x^2-y^2-3x+3y\)

Answer 1

`x^2-y^2-3x+3y`

`=(x^2-y^2) -(3x-3y)`

`=(x-y)(x+y)-3(x-y)`

`=(x-y)(x+y-3)`

Answer 2

\(x^2-y^2-3x+3y\)
\(=\left(x^2-y^2\right)-\left(3x-3y\right)\)
\(=\left(x+y\right)\left(x-y\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-3\right)\)