ĐKXĐ:\(-\sqrt{5}\)≤\(x\)≤\(\sqrt{5}\)
\(M^2=\left(2x+\sqrt{5-x^2}\right)^2\)≤\(\left(2^2+1^2\right)\left(x^2+5-x^2\right)=25\)
⇒ \(M\le5\)⇒ max\(M\)=5⇔ \(x=2\)
Mặt khác \(M=2x+\sqrt{5-x^2}\ge2\left(-\sqrt{5}\right)+0=-2\sqrt{5}\)
⇒ min\(M=-2\sqrt{5}\Leftrightarrow x=-\sqrt{5}\)