$Giải phương trình : \sqrt{2 - x^{2}} + \sqrt{2 - \frac{1}{x^{2}}} = 4 - \left ( x + \frac{1}{x} \right )$
<=>$\sqrt{2-x^2}+\sqrt{2-\frac{1}{x^2}}+(x+\frac{1}{x})=4 Ta có\left\{\begin{matrix} (1.\sqrt{2-x^2}+1.x)^2\leq4 & \\ (\sqrt{1-\frac{1}{x^2}}.1+\frac{1}{x}.1)^2\leq 4 \end{matrix}\right.=> Vt\leq 4=VP=>\left\{\begin{matrix} \sqrt{2-x^2}+x=2 & \\ \sqrt{2-\frac{1}{x^2}}+\frac{1}{x}=2 \end{matrix}\right.=>x=1$