Giải:
\(x^4-4x^2+4x+1\)
\(=\left(x^2\right)^2-\left(2x\right)^2+4x+1^2\)
\(=\left(x^2\right)^2-\left(2x\right)^2+4x-\left(-1\right)^2\)
\(=\left(x^2\right)^2-\left[\left(2x\right)^2-4x+\left(-1\right)^2\right]\)
\(=\left(x^2\right)^2-\left[2x-\left(-1\right)\right]^2\)
\(=\left(x^2\right)^2-\left(2x+1\right)^2\)
\(=\left[x^2-\left(2x+1\right)\right]\left[x^2+\left(2x+1\right)\right]\)
\(=\left(x^2-2x-1\right)\left(x^2+2x+1\right)\)
Chúc bạn học tốt!