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

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

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

Đặt \(y=x-1\) , ta có:

\(x^4 + yx^2-xy^2=0\)

\(⇔2y^2−x^2y−x^4=0(1) \)

Xem (1) là phương trình bậc 2 theo y . Do đó dấu bằng xảy ra

\(\left(1\right)\Leftrightarrow\left[{}\begin{matrix}y=x^2\\y=-\dfrac{x^2}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x-1=x^2\\x-1=-\dfrac{x^2}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\x^2+2x-2=0\end{matrix}\right.\) \(\Leftrightarrow x=-1\pm\sqrt{3}\)