\(\Leftrightarrow x^2-2+\dfrac{1}{x^2}+y^2-2+\dfrac{1}{y^2}=4-2-2\)
\(\Leftrightarrow\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
Với mọi x, y ta luôn có \(\left\{{}\begin{matrix}\left(x-\dfrac{1}{x}\right)^2\ge0\\\left(y-\dfrac{1}{y}\right)^2\ge0\end{matrix}\right.\)
=> \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2\ge0\)
mà \(\left(x-\dfrac{1}{x}\right)^2+\left(y-\dfrac{1}{y}\right)^2=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{1}{x}=0\\y-\dfrac{1}{y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2-1}{x}=0\\\dfrac{y^2-1}{y}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1;x=-1\\y=1;y=-1\end{matrix}\right.\)
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mk giải luôn đó nha
Giải:
Áp dụng BĐT AM-GM cho hai số dương, ta có:
\(x^2+\dfrac{1}{x^2}\ge2\sqrt{x^2.\dfrac{1}{x^2}}=2\)
\(y^2+\dfrac{1}{y^2}\ge2\sqrt{y^2.\dfrac{1}{y^2}}=2\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}+y^2+\dfrac{1}{y^2}\ge4\)
Dấu "=" xảy ra khi:
\(x=y=\pm1\)
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