\(x^3-4x^2+4x\)
\(=x^3-\left(4x^2-4x\right)\)
\(=x^3-4x\left(x-1\right)\)
\(=x\left[x^2-4\left(x-1\right)\right]\)
Phân tích thành nhân tử?
\(x^3-4x^2+4x\)
\(=x\left(x^2-4x+4\right)\)
\(=x\left(x^2-2.x.2+2^2\right)\)
\(=x\left(x-2\right)^2\)
\(=x\left(x-2\right)\left(x-2\right)\)
$x^3−4x^2+4x$
$=x^3−(4x^2−4x)$
$=x^3−4x(x−1)$
$=x[x^2−4(x−1)]$
$=x(x^2-4x+4)$