Question

Giải pt

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

Answer 1

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

Đặt \(x^2-x=t\)pt có dạng 

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

Với t = 1 => \(x^2-x-1=0\Leftrightarrow x^2-x+\dfrac{1}{4}-\dfrac{5}{4}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2-\dfrac{5}{4}=0\Leftrightarrow\left(x-\dfrac{1}{2}-\dfrac{\sqrt{5}}{2}\right)\left(x-\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)=0\)

\(\Leftrightarrow x=\dfrac{1+\sqrt{5}}{2};x=\dfrac{1-\sqrt{5}}{2}\)

Với t = -2 => \(x^2-x+2=0\Leftrightarrow x=-1;x=-2\)

Answer 2

   x ≓ -0.618033886