Question

giải phương trình:

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

Answer 1

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

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

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

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

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

Ta thấy: \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0,\forall x\in R\)

\(\Rightarrow x^2+2x-2=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}-1\\x=-\sqrt{3}-1\end{matrix}\right.\)