a) ĐKXĐ: \(-2\le x\le2\)
\(M^2=2-x+x+2+2\sqrt{\left(2-x\right)\left(x+2\right)}=4+2\sqrt{\left(2-x\right)\left(x+2\right)}\)
\(\ge4\)\(\Rightarrow M\ge2\) Vậy min M = 2\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)(thỏa mãn)
Mặt khác \(M^2=4+2\sqrt{\left(2-x\right)\left(2+x\right)}\le4+2-x+2+x=8\)
\(\Rightarrow M\le2\sqrt{2}\) Vậy max M = \(2\sqrt{2}\Leftrightarrow x=0\)(thỏa mãn)
Câu b tương tự nhé
ĐKXĐ: ...
\(M\ge\sqrt{2-x+x+2}=2\)
\(M_{min}=2\) khi \(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(M\le\sqrt{2\left(2-x+x+2\right)}=2\sqrt{2}\)
\(M_{max}=2\sqrt{2}\) khi \(2-x=x+2\Leftrightarrow x=0\)
\(N\ge\sqrt{x-3+5-x}=\sqrt{2}\)
\(N_{min}=\sqrt{2}\) khi \(\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
\(N\le\sqrt{2\left(x-3+5-x\right)}=2\)
\(N_{max}=2\) khi \(x-3=5-x\Leftrightarrow x=4\)
