Đáp án:
\[x = 1\]
Giải thích các bước giải:
ĐKXĐ: \(x \ge - 2\)
Ta có:
\(\begin{array}{l}
\sqrt {20{x^2} + 80x + 125} \le 2x + 1 + 4\sqrt {3x + 6} \\
\Leftrightarrow \left[ {\sqrt {20{x^2} + 80x + 125} - \left( {4x + 11} \right)} \right] + \left[ {\left( {2x + 10} \right) - 4\sqrt {3x + 6} } \right] \le 0\\
\Leftrightarrow \frac{{\left( {20{x^2} + 80x + 125} \right) - {{\left( {4x + 11} \right)}^2}}}{{\sqrt {20{x^2} + 80x + 125} + \left( {4x + 11} \right)}} + \frac{{{{\left( {2x + 10} \right)}^2} - 16\left( {3x + 6} \right)}}{{\left( {2x + 10} \right) + 4\sqrt {3x + 6} }} \le 0\\
\Leftrightarrow \frac{{4{x^2} - 8x + 4}}{{\sqrt {20{x^2} + 80x + 125} + \left( {4x + 11} \right)}} + \frac{{4{x^2} - 8x + 4}}{{\left( {2x + 10} \right) + 4\sqrt {3x + 6} }} \le 0\\
\Leftrightarrow 4{\left( {x - 1} \right)^2}\left( {\frac{1}{{\sqrt {20{x^2} + 80x + 125} + \left( {4x + 11} \right)}} + \frac{1}{{\left( {2x + 10} \right) + 4\sqrt {3x + 6} }}} \right) \le 0\\
x \ge - 2 \Rightarrow \frac{1}{{\sqrt {20{x^2} + 80x + 125} + \left( {4x + 11} \right)}} + \frac{1}{{\left( {2x + 10} \right) + 4\sqrt {3x + 6} }} > 0\\
\Rightarrow {\left( {x - 1} \right)^2} \le 0\\
\Rightarrow x - 1 = 0\\
\Leftrightarrow x = 1\left( {t/m} \right)
\end{array}\)
