Question

giải PT: \(x^4-2x^3+2x^2+4x-8=0\)

Answer 1

Lời giải:

Ta có:

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\left(1\right)\\x^2-2x+4=0\left(2\right)\end{matrix}\right.\)

(1) \(\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm \sqrt{2}\)

(2)\(\Leftrightarrow x^2-2x+4=0\Leftrightarrow (x-1)^2+3=0\)

(vô lý vì \((x-1)^2+3\geq 3>0\forall x\in\mathbb{R}\) )

Vậy \(x=\pm \sqrt{2}\)