ĐK: \(\left[{}\begin{matrix}x\le0\\x\ge1\end{matrix}\right.\)
\(x^4-2x^3+x-\sqrt{2\left(x^2-x\right)}=0\)
\(\Leftrightarrow x\left(x^3-2x^2+1\right)-\sqrt{2\left(x^2-x\right)}=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2-x-1\right)-\sqrt{2\left(x^2-x\right)}=0\)
\(\Leftrightarrow\left(x^2-x\right)\left(x^2-x-1\right)-\sqrt{2\left(x^2-x\right)}=0\)
Đặt \(\sqrt{x^2-x}=a\)
\(pt\Leftrightarrow a^2\left(a^2-1\right)-a\sqrt{2}=0\)
\(\Leftrightarrow a^4-a^2-a\sqrt{2}=0\)
\(\Leftrightarrow a\left(a^3-a-\sqrt{2}\right)=0\)
\(\Leftrightarrow a\left[a^2\left(a-\sqrt{2}\right)+a\sqrt{2}\left(a-\sqrt{2}\right)+\left(a-\sqrt{2}\right)\right]=0\)
\(\Leftrightarrow a\left(a-\sqrt{2}\right)\left(a^2+a\sqrt{2}+1\right)=0\)
Vì \(a^2+a\sqrt{2}+1>0\forall a\)( tự chứng minh )
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x}=0\\\sqrt{x^2-x}=\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=0\\x^2-x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\left(x-2\right)\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\\x=-1\end{matrix}\right.\)( thỏa )
Vậy....
