[tex]\frac{1}{x}+\frac{1}{\sqrt{2-x^2}}=2[/tex]
ĐKXĐ:...
Đặt [tex]\sqrt{2-x^{2}}=y(y> 0)[/tex]
Khi đó, ta có:
$\left\{\begin{matrix} \frac{1}{x}+\frac{1}{y}=2\\x^{2}+y^{2}=2 \end{matrix}\right. $
$\Leftrightarrow \left\{\begin{matrix} x+y=2xy\\(x+y)^{2}-2xy=2 \end{matrix}\right. $
$\Leftrightarrow \left\{\begin{matrix} x+y=2xy\\4x^{2}y^{2}-2xy-2=0 \end{matrix}\right. $
$\Leftrightarrow \left\{\begin{matrix} x+y=2xy\\2(xy-1)(2xy+1)=0 \end{matrix}\right. $
$\Leftrightarrow \left\{\begin{matrix} x+y=2\\ xy=1 \end{matrix}\right.$ hoặc [tex]\left\{\begin{matrix} x+y=-1\\ xy=\frac{-1}{2} \end{matrix}\right. \Leftrightarrow ....[/tex]
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