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=> \(\left(x^2-2+\frac{1}{x^2}\right)+\left(y^2-2+\frac{1}{y^2}\right)=0\)
=> \(\left(x^2-2x.\frac{1}{x}+\frac{1}{x^2}\right)+\left(y^2-2.y.\frac{1}{y}+\frac{1}{y^2}\right)=0\)
=> \(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
<=> \(x-\frac{1}{x}=0;y-\frac{1}{y}=0\)
=> \(x^2=1;y^2=1\)
=> x = 1 hoặc -1
y = 1 hoặc -1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4\)
\(\Leftrightarrow\)\(\left(x^2+\frac{1}{x^2}-2\right)+\left(y^2+\frac{1}{y^2}-2\right)=0\)
\(\Leftrightarrow\)\(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}x-\frac{1}{x}=0\\y-\frac{1}{y}=0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x^2=1\\y^2=1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x=\pm1\\y=\pm1\end{cases}}\)
\(^{x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=4}\)
\(\Leftrightarrow\left(x^2+\frac{1}{x^2}\right)+\left(y^2+\frac{1}{y^2}\right)-2-2=0\)
\(\Leftrightarrow\left(x^2-2.x.\frac{1}{x}+\frac{1}{x^2}\right)+\left(y^2-2.y.\frac{1}{y}+\frac{1}{y^2}\right)=0\)
\(\Leftrightarrow\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0\)
Mặt khác: \(\left(x-\frac{1}{x}\right)^2\ge0\)\(\forall\)x\(\ne\)0
\(\left(y-\frac{1}{y}\right)^2\ge0\)\(\forall\)y \(\ne\)0
Từ hai điều trên \(\Rightarrow\)\(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2\ge0\)\(\forall\)x,y \(\ne\)0
Dấu "=" xảy ra
![](https://rs.olm.vn/images/avt/0.png?1311)
\(pt\Leftrightarrow\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{4}-\frac{x^2+y^2+z^2}{5}=0\)
\(\Leftrightarrow\left(\frac{x^2}{2}-\frac{x^2}{5}\right)+\left(\frac{y^2}{3}-\frac{y^2}{5}\right)+\left(\frac{z^2}{4}-\frac{z^2}{5}\right)=0\)
\(\Leftrightarrow\frac{3}{10}x^2+\frac{2}{15}y^2+\frac{1}{20}z^2=0\)
Ta thấy \(VT\ge0\forall x;y;z\) nên để dấu "=" xảy ra \(\Leftrightarrow x=y=z=0\)
rút gọn P=2/x-(x2/(x2-xy)+(x2-y2)/xy-y2/(y2-xy)):(x2-xy+y2)/(x-y)
r tìm gt P với |2x-1|=1 ; |y+1|=1/2
![](https://rs.olm.vn/images/avt/0.png?1311)
=> \(\left(x-\frac{1}{x}\right)^2+\left(y-\frac{1}{y}\right)^2=0=>\)
\(\hept{\begin{cases}y-\frac{1}{y}=0\orbr{\begin{cases}y=1\\y-1\end{cases}}\\x-\frac{1}{x}=0\orbr{\begin{cases}x=1\\x=-1\end{cases}}\end{cases}}\)
=>
x=+-1
y=+-1