Lời giải:
ĐKXĐ: \(-2\leq x\leq 2\)
Đặt \(\sqrt{x+2}=a; \sqrt{2-x}=b( a,b\geq 0)\)
\(\Rightarrow \)\(a^2+b^2=4\)
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Ta có: \(y=a+b+2ab\)
Tìm min:
\(y=\sqrt{(a+b)^2}+2ab=\sqrt{a^2+b^2+2ab}+2ab=\sqrt{4+2ab}+ab\)
Vì \(a,b\in [0;2]\Rightarrow ab\geq 0\)
\(\Rightarrow y\geq \sqrt{4+0}+0\Leftrightarrow y\geq 2\)
Vậy \(y_{\min}=2\Leftrightarrow ab=0\Leftrightarrow x=\pm 2\)
Tìm max:
Áp dụng BĐT Am-Gm:
\(ab\leq \frac{(a+b)^2}{4}\Rightarrow y\leq a+b+\frac{(a+b)^2}{2}\) (1)
Tiếp tục áp dụng AM-GM:
\(a^2+b^2\geq 2ab\Rightarrow 2(a^2+b^2)\geq (a+b)^2\)
\(\Leftrightarrow (a+b)^2\leq 8\Rightarrow a+b\leq 2\sqrt{2}\) (2)
Từ \((1),(2)\Rightarrow y\leq 2\sqrt{2}+\frac{8}{2}=4+2\sqrt{2}\)
Vậy \(y_{\max}=4+2\sqrt{2}\). Dấu bằng xảy ra khi \(a=b\Leftrightarrow \sqrt{2+x}=\sqrt{2-x}\Leftrightarrow x=0\)
\(y_{\max}+y_{\min}=2+4+2\sqrt{2}=6+2\sqrt{2}\)
