Question

Giải phương trình :

\(x^6-6x^4-64x^3+12x^2-8=0\)

Answer 1

\(x^6-6x^4-64x^3+12x^2-8=0\)

\(\Leftrightarrow\left(x^2-4x-2\right)\left(x^4+4x^3+12x^2-8x+4\right)=0\)

\(\Leftrightarrow\left(x^2-4x-2\right)\left[\left(x^4+4x^3+4x^2\right)+\left(8x^2-8x+\frac{8}{4}\right)+2\right]=0\)

\(\Leftrightarrow\left(x^2-4x-2\right)\left[\left(x^2+2x\right)^2+8\left(x-\frac{1}{2}\right)^2+2\right]=0\)

\(\Leftrightarrow x^2-4x-2=0\)

\(\Leftrightarrow x=2\pm\sqrt{6}\)