Ta có:\(2x^2+\dfrac{1}{x^2}+\dfrac{y^2}{4}=4\)
<=>\(0=\left(x^2-2+\dfrac{1}{x^2}\right)+\left(x^2+\dfrac{y^2}{4}\right)-2=\left(x-\dfrac{1}{x}\right)^2+\left(x^2+xy+\dfrac{y^2}{4}\right)-xy-2=\left(x-\dfrac{1}{x}\right)^2+\left(x+\dfrac{y}{2}\right)^2-xy-2\)
<=>xy+2=\(\left(x-\dfrac{1}{x}\right)^2+\left(x+\dfrac{y}{2}\right)^2\)
Do \(\left(x-\dfrac{1}{x}\right)^2\ge0\)
\(\left(x+\dfrac{y}{2}\right)^2\ge0\)
=>xy+2\(\ge\)0
=>P=xy\(\ge\)-2
=>GTNN của P=-2 khi và chỉ khi \(\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\x+\dfrac{y}{2}=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
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