Ta có: \(x^2-x+1=x^2-2.x.\frac{1}{2}+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\in R\)
Thấy \(x^8\ge0;x^5< x^8\Rightarrow x^8-x^5\ge0\)
\(\Rightarrow x^8-x^5+x^2-x+1>0\forall x\in R.\)(đpcm)
x = 0
co dung KO thi nho ket ban voi minh nha ^_^