\(x^2=x^5\left(đk:x\ge0\right)\)
\(\Leftrightarrow x^5-x^2=0\Leftrightarrow x^2\left(x^3-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmđk\right)\\x^3-1=0\left(1\right)\end{matrix}\right.\)
Từ (1) --> \(\left(x-1\right)\left(x^2+x+1\right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tmđk\right)\) vì: \(x^2+x+1=\dfrac{1}{2}\left(\left(x+1\right)^2+x^2+1\right)>0\)
Vậy x = 0 ; x = 1
