\(x^4-x^3+2x^2-x+1=0\)
\(\Rightarrow\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)=0\)
\(\Rightarrow x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)=0\)
\(\Rightarrow\left(x^2+1\right)\left(x^2-x+1\right)=0\)
Mà \(\hept{\begin{cases}x^2+1>0\forall x\\x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\forall x\end{cases}\Rightarrow\left(x^2+1\right)\left(x^2-x+1\right)>0\forall x}\)
Vậy ko tồn tại x thỏa mãn \(x^4-x^3+2x^2-x+1=0\)
\(x^4-x^3+2x^2-x+1=x^4-x^3+x^2+x^2-x+1\)
\(=x^2.\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2+1\right).\left(x^2-x+1\right)\)
vì (x2+1) \(\ge1\)
và \(x^2\ge x\Rightarrow x^2-x+1\ge1\)
=> \(\left(x^2+1\right).\left(x^2-x+1\right)\ge1\Rightarrowđpcm\)
đoạn này t sai r :(
\(x^2-x+1=x^2-\frac{2x.1}{2}+\frac{1}{2^2}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
=> \(\left(x^2+1\right).\left(x^2-x+1\right)\ge\frac{3}{4}\)=> đpcm