Ta có:
\(M=\frac{4x^2}{x^4+1}=\frac{4x^2}{x^4+1}-2+2=\frac{-2x^4+4x-2}{x^4+1}+2\)
\(M=\frac{-2\left(x^2-1\right)^2}{x^4+1}+2\le2\)
Dấu "=" xảy ra khi \(x^2-1=0\Leftrightarrow x\in\left\{\pm1\right\}\)
Vậy \(x\in\left\{\pm1\right\}\Leftrightarrow maxM=2\)