Ta có:
\(x^2+x+1=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x^2+\dfrac{1}{2}x\right)+\left(\dfrac{1}{2}x+\dfrac{1}{4}\right)+\dfrac{3}{4}\)
\(=x.\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}.\left(x+\dfrac{1}{2}\right)+\dfrac{3}{2}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Với mọi giá trị của \(x\in R\) ta có:
\(\left(x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\) với mọi giá trị của \(x\in R\)
Vậy \(x^2+x+1>0\) (đpcm)
Chúc bạn học tốt!!!
Ta có: \(x^2+x+1=x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
\(\Rightarrowđpcm\)
Ta có: \(x^2+x+1=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+1\)
\(=x^2+\dfrac{1}{2}x+\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=x\left(x+\dfrac{1}{2}\right)+\dfrac{1}{2}\left(x+\dfrac{1}{2}\right)+\dfrac{3}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x+\dfrac{1}{2}\right)^2\ge0\) \(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\forall x\)
\(\rightarrowđpcm.\)
