$\sqrt{x^2 -\dfrac{7}{x^2}} + \sqrt{x - \dfrac{7}{x^2}}= 0$ $(*)$
$ĐK: \, \begin{cases}x \ne 0\\x^2 -\dfrac{7}{x^2}\geq 0\\x -\dfrac{7}{x^2}\geq 0\end{cases} \Leftrightarrow x \geq \sqrt[3]{7}$
Ta có:
$\sqrt{x^2 -\dfrac{7}{x^2}} \geq 0$
$\sqrt{x - \dfrac{7}{x^2}} \geq 0$
$\Rightarrow \sqrt{x^2 -\dfrac{7}{x^2}} + \sqrt{x - \dfrac{7}{x^2}} \geq 0$
Dấu = xảy ra $\Leftrightarrow \begin{cases}\sqrt{x^2 -\dfrac{7}{x^2}} = 0\\\sqrt{x - \dfrac{7}{x^2}} =0\end{cases}$
$\Leftrightarrow \begin{cases}x^2 -\dfrac{7}{x^2} =0\\x - \dfrac{7}{x^2}=0\end{cases}$
$\Leftrightarrow x^2 = x$
$\Leftrightarrow \left[\begin{array}{l}x = 0\\x = 1\end{array}\right.\qquad \text{(không thoả ĐK)}$
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