\(x^2+x+3\)
\(=x^2+x+\frac{1}{4}+\frac{11}{4}\)
\(=x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}+\frac{11}{4}\)
\(=x\left(x+\frac{1}{2}\right)+\frac{1}{2}\left(x+\frac{1}{2}\right)+\frac{11}{4}\)
\(=\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\frac{11}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Vậy ...
\(x^2+x+3=x^2+\frac{1}{2}.2.x+\frac{1}{4}+\frac{11}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\) .......................... Đúng 100% ...........................
\(\frac{11}{4}>0\) ................................. Tk cho mình nha! ...................................
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{11}{4}>0\)
\(\Rightarrow x^2+x+3>0\)
Hay \(x^2+x+3\)luôn dương với mọi x
\(x^2+x+3=\left(x^2+x+\frac{1}{4}\right)+\frac{11}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{11}{4}>0\forall x\)
Vậy ............
