Đáp án:
\[x = 2\]
Giải thích các bước giải:
ĐKXĐ: \(1 \le x \le 11\)
Ta có:
\(\begin{array}{l}
\sqrt {x - 1} - x\sqrt {11 - x} = {x^2} - 4x - 1\\
\Leftrightarrow \left( {\sqrt {x - 1} - 1} \right) - \left( {x\sqrt {11 - x} - 3x} \right) = {x^2} - x - 2\\
\Leftrightarrow \left( {\sqrt {x - 1} - 1} \right) - x\left( {\sqrt {11 - x} - 3} \right) = {x^2} - x - 2\\
\Leftrightarrow \frac{{x - 1 - 1}}{{\sqrt {x - 1} + 1}} - x.\frac{{11 - x - 9}}{{\sqrt {11 - x} + 3}} = \left( {x - 2} \right)\left( {x + 1} \right)\\
\Leftrightarrow \frac{{x - 2}}{{\sqrt {x - 1} + 1}} + x.\frac{{x - 2}}{{\sqrt {11 - x} + 3}} = \left( {x - 1} \right)\left( {x + 1} \right)\\
\Leftrightarrow \left( {x - 2} \right)\left[ {\frac{1}{{\sqrt {x - 1} + 1}} + \frac{x}{{\sqrt {11 - x} + 3}} - \left( {x + 1} \right)} \right] = 0\\
\frac{1}{{\sqrt {x - 1} + 1}} - 1 = \frac{{1 - \sqrt {x - 1} - 1}}{{\sqrt {x - 1} + 1}} = \frac{{ - \sqrt {x - 1} }}{{\sqrt {x - 1} + 1}} < 0,\,\,\,\,\forall x \in \left[ {1;11} \right]\\
\frac{x}{{\sqrt {11 - x} + 3}} - x = \frac{{x - x\sqrt {11 - x} - 3x}}{{\sqrt {11 - x} + 3}} = \frac{{ - 2x - x\sqrt {11 - x} }}{{\sqrt {11 - x} + 3}} < 0,\,\,\,\forall x \in \left[ {1;11} \right]\\
\Rightarrow \frac{1}{{\sqrt {x - 1} + 1}} + \frac{x}{{\sqrt {11 - x} + 3}} - \left( {x + 1} \right) < 0,\,\,\,\forall x \in \left[ {1;11} \right]\\
\Rightarrow x - 2 = 0\\
\Leftrightarrow x = 2
\end{array}\)
