Có 3 người nhé bạn 

k cho mình nha

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

Vậy ta chọn câu D

a)\(x^3+2x^2y+xy^2-9x\)

\(=x\left(x^2+2xy+y^2-9\right)\)

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

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

b)\(2x-2y-x^2+2xy+y^2\)

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

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

\(=\left(x-y\right)\left[2-\left(x-y\right)\right]\)

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

c)\(x^4-2x^2\)

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

\(=x^2\left[x^2-\left(\sqrt{2}\right)^2\right]\)

\(=x^2\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\)

B

nghiệm là j? mk chưa hok đến!

\(x-\frac{1}{x-1}=1-\frac{1}{x-1}\)

\(\Leftrightarrow\frac{x\left(x-1\right)}{x-1}-\frac{1}{x-1}=\frac{x-1}{x-1}-\frac{1}{x-1}\)

\(\Leftrightarrow\frac{x^2-x-1}{x-1}=\frac{x-1-1}{x-1}\Leftrightarrow x^2-x-1=x-2\Leftrightarrow x=1\)