Theo bài ra ta có:
\(x^2-x+1=x^2-\dfrac{1}{2}.2.x+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
Dấu \("="\) xảy ra khi \(x-\dfrac{1}{2}=0=>x=\dfrac{1}{2}\)
Mà : \(\dfrac{3}{4}>0\)
\(=>x^2-x+1>0\)
CHÚC BẠN HỌC TỐT..........
\(x^2-x+1\\ = x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\\= \left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Ta có:
\(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\\ \Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
Vậy \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\left(đpcm\right)\)
Z 
