Question

cm bdt \(x^2+x\sqrt{2}+1>0\)

Answer 1

\(x^2+x\sqrt{2}+1>0\)

\(\Leftrightarrow\left(x+\frac{1}{\sqrt{2}}\right)^2+\frac{1}{2}>0\)

\(\Leftrightarrow\left(x+\frac{1}{\sqrt{2}}\right)^2>-\frac{1}{2}\)

=> đpcm

Answer 2

\(x^2+x\sqrt{2}+1=x^2+2.x.\frac{\sqrt{2}}{2}+\left(\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}=x^2+2.x.\frac{\sqrt{2}}{2}+\frac{1}{2}+\frac{1}{2}\)

                                                              \(=\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}\)

Vì \(\left(x+\frac{\sqrt{2}}{2}\right)^2\ge0\left(\forall x\right)\)

Suy ra: \(\left(x+\frac{\sqrt{2}}{2}\right)^2+\frac{1}{2}\ge\frac{1}{2}>0\)

Vậy \(x^2+x\sqrt{2}+1>0\)