\(x\ge0\)
\(\Leftrightarrow-x+\sqrt{x}+1=\sqrt{2\left(x^2-x+1\right)}\)
Do \(VP>0\Rightarrow VT>0\Rightarrow0\le x\le\frac{3+\sqrt{5}}{2}\)
Đặt \(\sqrt{x}=a\ge0\Rightarrow-a^2+a+1=\sqrt{2\left(a^4-a^2+1\right)}\)
\(\Leftrightarrow\left(-a^2+a+1\right)^2=2\left(a^4-a^2+1\right)\)
\(\Leftrightarrow a^4+a^2+1-2a^3-2a^2+2a=2a^4-2a^2+2\)
\(\Leftrightarrow a^4+2a^3-a^2-2a+1=0\)
Do \(a=0\) không phải nghiệm, chia 2 vế cho \(a^2\)
\(\Leftrightarrow a^2+\frac{1}{a^2}+2\left(a-\frac{1}{a}\right)-1=0\)
Đặt \(a-\frac{1}{a}=t\Rightarrow a^2+\frac{1}{a^2}=t^2+2\)
\(\Rightarrow t^2+2+2t-1=0\Leftrightarrow t^2+2t+1=0\Rightarrow t=-1\)
\(\Rightarrow a-\frac{1}{a}=-1\Leftrightarrow a^2+a-1=0\Rightarrow\left[{}\begin{matrix}a=\frac{\sqrt{5}-1}{2}\\a=\frac{-\sqrt{5}-1}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=a^2=\left(\frac{\sqrt{5}-1}{2}\right)^2=\frac{3-\sqrt{5}}{2}\)
