√(30-5/x^2)+√(6x^2-5/x^2)=6x^2 mọi người giúp mk vs khó quá cảm mơn nhiều 01/09/2021 Bởi Valentina √(30-5/x^2)+√(6x^2-5/x^2)=6x^2 mọi người giúp mk vs khó quá cảm mơn nhiều
Đáp án: \[x = \pm 1\] Giải thích các bước giải: \[\begin{array}{l}\sqrt {30 – \frac{5}{{{x^2}}}} + \sqrt {6{x^2} – \frac{5}{{{x^2}}}} = 6{x^2}\\dk:\left[ {\begin{array}{*{20}{c}}{x \ge \sqrt[4]{{\frac{5}{6}}}}\\{x \le – \sqrt[4]{{\frac{5}{6}}}}\end{array}} \right.\\ \Leftrightarrow \sqrt {30 – \frac{5}{{{x^2}}}} – 5 + \sqrt {6{x^2} – \frac{5}{{{x^2}}}} – 1 = 6{x^2} – 6\\ \Leftrightarrow \frac{{30 – \frac{5}{{{x^2}}} – 25}}{{\sqrt {30 – \frac{5}{{{x^2}}}} + 5}} + \frac{{6{x^2} – \frac{5}{{{x^2}}} – 1}}{{\sqrt {6{x^2} – \frac{5}{{{x^2}}}} + 1}} = 6({x^2} – 1)\\ \Leftrightarrow \frac{{\frac{5}{{{x^2}}}({x^2} – 1)}}{{\sqrt {30 – \frac{5}{{{x^2}}}} }} + \frac{{({x^2} – 1)(6{x^2} + 5)}}{{\sqrt {6{x^2} – \frac{5}{{{x^2}}}} + 1}} = 6({x^2} – 1)\\ \Leftrightarrow ({x^2} – 1)(\frac{5}{{{x^2}\sqrt {30 – \frac{5}{{{x^2}}}} }} + \frac{{6{x^2} + 5}}{{\sqrt {6{x^2} – \frac{5}{{{x^2}}} + 1} }} – 6) = 0\\ \Leftrightarrow {x^2} – 1 = 0 \Leftrightarrow x = \pm 1(tm)\end{array}\] Bình luận
Đáp án:
\[x = \pm 1\]
Giải thích các bước giải:
\[\begin{array}{l}
\sqrt {30 – \frac{5}{{{x^2}}}} + \sqrt {6{x^2} – \frac{5}{{{x^2}}}} = 6{x^2}\\
dk:\left[ {\begin{array}{*{20}{c}}
{x \ge \sqrt[4]{{\frac{5}{6}}}}\\
{x \le – \sqrt[4]{{\frac{5}{6}}}}
\end{array}} \right.\\
\Leftrightarrow \sqrt {30 – \frac{5}{{{x^2}}}} – 5 + \sqrt {6{x^2} – \frac{5}{{{x^2}}}} – 1 = 6{x^2} – 6\\
\Leftrightarrow \frac{{30 – \frac{5}{{{x^2}}} – 25}}{{\sqrt {30 – \frac{5}{{{x^2}}}} + 5}} + \frac{{6{x^2} – \frac{5}{{{x^2}}} – 1}}{{\sqrt {6{x^2} – \frac{5}{{{x^2}}}} + 1}} = 6({x^2} – 1)\\
\Leftrightarrow \frac{{\frac{5}{{{x^2}}}({x^2} – 1)}}{{\sqrt {30 – \frac{5}{{{x^2}}}} }} + \frac{{({x^2} – 1)(6{x^2} + 5)}}{{\sqrt {6{x^2} – \frac{5}{{{x^2}}}} + 1}} = 6({x^2} – 1)\\
\Leftrightarrow ({x^2} – 1)(\frac{5}{{{x^2}\sqrt {30 – \frac{5}{{{x^2}}}} }} + \frac{{6{x^2} + 5}}{{\sqrt {6{x^2} – \frac{5}{{{x^2}}} + 1} }} – 6) = 0\\
\Leftrightarrow {x^2} – 1 = 0 \Leftrightarrow x = \pm 1(tm)
\end{array}\]