\(dk:-\sqrt{2}\le x\le\sqrt{2}\)(*)
\(\left(A-3x\right)=\sqrt{2-x^2}\)
\(\Leftrightarrow a^2-6ax+9x^2=2-x^2\)
\(10x^2-6ax+a^2-2=0\)(**)
Giá trị (a) (**) có nghiệm thỏa mãn (*)
(**)\(\Leftrightarrow\left(x^2-2.\frac{3a}{10}x+\frac{9a^2}{100}\right)=\frac{20-a^2}{100}\)\(\Leftrightarrow\left(x-\frac{3a}{10}\right)^2=\frac{20-a^2}{100}\Rightarrow20-a^2\ge0\Rightarrow!a!\le2\sqrt{5}\)
\(a=2\sqrt{5}\Rightarrow x=\frac{6\sqrt{5}}{10}=\frac{3\sqrt{5}}{5}< \sqrt{2}\)(nhạn)
Kết luận: GTLN của A là \(A_{max}=2\sqrt{5}\) tại \(x=\frac{3\sqrt{5}}{5}\)
\(A=3x+\sqrt{2-x^2}\)
\(\Leftrightarrow\frac{\sqrt{10}A}{2}=\frac{3\sqrt{10}x}{2}+\frac{\sqrt{10}\sqrt{2-x^2}}{2}\)
\(\le\sqrt{\left(\frac{45}{2}+\frac{5}{2}\right)\left(x^2+2-x^2\right)}=\sqrt{25.2}=5\sqrt{2}\)
\(\Rightarrow1A\le\frac{5\sqrt{2}.2}{\sqrt{10}}=2\sqrt{5}\)
Vậy GTLN là A = \(2\sqrt{5}\)khi x = \(\frac{3}{\sqrt{5}}\)
