a ) Để A = \(\sqrt{10}\Leftrightarrow\sqrt{3-x}+\sqrt{3+x}=\sqrt{10}\)
\(\Leftrightarrow3-x+3+x+2\sqrt{9-x^2}=10\)
\(\Leftrightarrow6+2\sqrt{9-x^2}=10\)
\(\Leftrightarrow2\sqrt{9-x^2}=4\)
\(\Leftrightarrow\sqrt{9-x^2}=2\)
\(\Leftrightarrow9-x^2=4\)
\(\Leftrightarrow-x^2=-5\)
\(\Leftrightarrow x^2=5\)
\(\Leftrightarrow x=\pm\sqrt{5}\)
:v Fan Hương Tràm
b) Áp dụng BĐT Bunyakovsky, ta có:
\(\left(3-x+3+x\right)2\ge\left(\sqrt{3-x}+\sqrt{3+x}\right)^2\)
\(\Leftrightarrow12\ge A^2\)
\(\Leftrightarrow A\le\sqrt{12}=2\sqrt{3}\)
Max \(A=2\sqrt{3}\Leftrightarrow x=0\)
Lại có: \(A^2=3-x+3+x+2.\sqrt{3-x}.\sqrt{3+x}\ge3-x+3+x=6\)
\(\Leftrightarrow A\ge\sqrt{6}\)
Min \(A=\sqrt{6}\Leftrightarrow\)\(\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
