Question

Chứng minh rằng : x² - x + 1 > 0 với mọi x.

Answer 1

Ta có:\(x^2-x+1=x^2-2.\dfrac{1}{2}.x+\dfrac{1}{4}+\left(1-\dfrac{1}{4}\right)=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge0\forall x\)

Do \(\left(x-\dfrac{1}{2}\right)^2\ge0\forall x\)

Vậy....

Answer 2

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

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

Với mọi giá trị của x ta có:

\(\left(x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

=> đpcm