Khai triển :
\(P=\frac{20-x^2}{5+x^2}=\frac{-x^2-5+25}{x^2+5}=\frac{25}{x^2+5}-1\)
Ta có : \(x^2\ge0\left(\forall x\right)\)
\(\Leftrightarrow x^2+5\ge5\)
\(\Leftrightarrow\frac{25}{x^2+5}-1\le4\)
Dấu " = " xảy ra <=> x = 0
Vậy MaxP=4 khi x = 0