Đáp án:
\[x = \pm 1\]
Giải thích các bước giải:
\[\begin{array}{l}
\sqrt {30 - \frac{5}{{{x^2}}}} + \sqrt {6{x^2} - \frac{5}{{{x^2}}}} = 6{x^2}\\
dk:\left[ {\begin{array}{*{20}{c}}
{x \ge \sqrt[4]{{\frac{5}{6}}}}\\
{x \le - \sqrt[4]{{\frac{5}{6}}}}
\end{array}} \right.\\
\Leftrightarrow \sqrt {30 - \frac{5}{{{x^2}}}} - 5 + \sqrt {6{x^2} - \frac{5}{{{x^2}}}} - 1 = 6{x^2} - 6\\
\Leftrightarrow \frac{{30 - \frac{5}{{{x^2}}} - 25}}{{\sqrt {30 - \frac{5}{{{x^2}}}} + 5}} + \frac{{6{x^2} - \frac{5}{{{x^2}}} - 1}}{{\sqrt {6{x^2} - \frac{5}{{{x^2}}}} + 1}} = 6({x^2} - 1)\\
\Leftrightarrow \frac{{\frac{5}{{{x^2}}}({x^2} - 1)}}{{\sqrt {30 - \frac{5}{{{x^2}}}} }} + \frac{{({x^2} - 1)(6{x^2} + 5)}}{{\sqrt {6{x^2} - \frac{5}{{{x^2}}}} + 1}} = 6({x^2} - 1)\\
\Leftrightarrow ({x^2} - 1)(\frac{5}{{{x^2}\sqrt {30 - \frac{5}{{{x^2}}}} }} + \frac{{6{x^2} + 5}}{{\sqrt {6{x^2} - \frac{5}{{{x^2}}} + 1} }} - 6) = 0\\
\Leftrightarrow {x^2} - 1 = 0 \Leftrightarrow x = \pm 1(tm)
\end{array}\]
