Question

Giải phương trình: \(x^4+x^2+6x-8=0\)

Answer 1

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

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

\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)+2x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2+2x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x^3+2x^2-x^2-2x+4x+8\right]=0\)

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

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x+4\right)=0\)

Mà \(x^2-2x+4=x^2-2x+1+3=\left(x-1\right)^2+3>0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy........