Vì \(|x^2+\frac{1}{10}|\ge0\)\(\forall x\)
\(\Rightarrow\frac{9}{10}+|x^2+\frac{1}{10}|\ge\frac{9}{10}\)\(\forall x\)
hay \(C\ge\frac{9}{10}\)
\(\Rightarrow maxC=\frac{9}{10}\Leftrightarrow x^2+\frac{1}{10}=0\)
\(\Leftrightarrow x^2=\frac{-1}{10}\)
\(\Leftrightarrow x=\sqrt{\frac{-1}{10}}\)hoặc \(x=-\sqrt{\frac{-1}{10}}\)( vô lý )
Vậy \(x\in\varnothing\)