Ta có:\(x^3-3x^2+2\)
\(=x^3-x^2-2x^2+2\)
\(=\left(x^3-x^2\right)-\left(2x^2-2\right)\)
\(=x^2\left(x-1\right)-2\left(x^2-1\right)\)
\(=x^2\left(x-1\right)-2\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left[x^2-2.\left(x+1\right)\right]\)
\(=\left(x-1\right)\left(x^2-2x-2\right)\)
chúc bạn hok tốt !!!
\(x^3-3x^2+2\)
=\(x^3-x^2-2x^2+2\)
=[\(x^3-x^2\)]-[\(2x^2-2\)]
=\(x^2\)[x-1] - 2[x-1]
=[x-1][\(x^2-2\)]