Question

Cho hai số x,y thỏa mãn \(^{2x^2+\frac{1}{x^2}+\frac{y^2}{4}=4}\)

Answer 1

Đề thiếu bạn gì ơi!

Answer 2

\(2=x^2+\frac{1}{x^2}-2+x^2+\frac{y^2}{4}=\left(x-\frac{1}{x}\right)^2+x^2+\frac{y^2}{4}\ge x^2+\frac{y^2}{4}\)

\(\Rightarrow2\ge x^2+\frac{y^2}{4}\ge2\sqrt{\frac{x^2y^2}{4}}=\left|xy\right|\)

\(\Rightarrow-2\le xy\le2\)

\(\Rightarrow xy_{min}=-2\) khi \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

\(xy_{max}=2\) khi \(\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=-1\\y=-2\end{matrix}\right.\)