`(x^2-x)(x^2-x+1)=6`
Đặt `a=x^2-x(x>=-1/4)`
`pt<=>a(a+1)=6`
`<=>a^2+a-6=0`
`Delta=1+24=25`
`=>a_1=-2(l),a_2=1(tm)`
`<=>x^2-x=1`
`<=>x^2-x-1=0`
`Delta=1+4=5`
`=>x_{12}=(+-sqrt5+1)/2`
\(\left(x^2-x\right)\left(x^2-x+1\right)=6\)
\(\Leftrightarrow\left(x^2-x\right)^2+\left(x^2-x\right)-6=0\)
\(\Leftrightarrow\left[\left(x^2-x\right)-2\right]\left[\left(x^2-x\right)+3\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x=2\\x^2-x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x-2=0\\x^2-x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)\left(x+1\right)=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow\)\(\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
