Đặt \(t=\left(x^2-x+1\right)^2\) thì pt đã cho trở thành :
\(t^2+4x^2t=5x^4\)
\(\Leftrightarrow t^2+4x^2t-5x^4=0\)
\(\Leftrightarrow\left(t-x^2\right)\left(t+5x^2\right)=0\)
\(\Leftrightarrow t-x^2=0\) ( do \(t+5x^2=\left[\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\right]^2+5x^2>0\forall x\))
\(\Leftrightarrow\left(x^2-x+1\right)-x^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2=0\) ( do \(x^2+1>0\forall x\) )
\(\Leftrightarrow x=1\left(TM\right)\)