Question

Chứng minh: x²-x+1 >0 \(\forall\)x

Answer 1

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..........

Answer 2

\(x^2-x+1\\ = x^2-2\cdot\dfrac{1}{2}\cdot x+\dfrac{1}{4}+\dfrac{3}{4}\\= \left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

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 \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\forall x\left(đpcm\right)\)