Đáp án: a) $S = \left\{ {\dfrac{{11 - 3\sqrt 5 }}{2};3} \right\}$
b) $S = \left\{ {\dfrac{1}{2}} \right\}$
Giải thích các bước giải:
a) ĐKXĐ: $x\ge 2$
Ta có:
$\begin{array}{l}
3\left( {2 + \sqrt {x - 2} } \right) = 2x + \sqrt {x + 6} \\
\Leftrightarrow 3\sqrt {x - 2} - \sqrt {x + 6} + 6 - 2x = 0\\
\Leftrightarrow \sqrt {9x - 18} - \sqrt {x + 6} + 6 - 2x = 0\\
\Leftrightarrow \dfrac{{9x - 18 - x - 6}}{{\sqrt {9x - 18} + \sqrt {x + 6} }} - 2\left( {x - 3} \right) = 0\\
\Leftrightarrow \dfrac{{8\left( {x - 3} \right)}}{{\sqrt {9x - 18} + \sqrt {x + 6} }} - 2\left( {x - 3} \right) = 0\\
\Leftrightarrow \left( {x - 3} \right)\left( {\dfrac{4}{{\sqrt {9x - 18} + \sqrt {x + 6} }} - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 0\\
\dfrac{4}{{\sqrt {9x - 18} + \sqrt {x + 6} }} - 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3(tm)\\
3\sqrt {x - 2} + \sqrt {x + 6} = 4
\end{array} \right.\\
+ )3\sqrt {x - 2} + \sqrt {x + 6} = 4\\
\Leftrightarrow 9\left( {x - 2} \right) + x + 6 + 6.\sqrt {x - 2} .\sqrt {x + 6} = 16\\
\Leftrightarrow 3\sqrt {\left( {x - 2} \right)\left( {x + 6} \right)} = 14 - 5x\\
\Leftrightarrow \left\{ \begin{array}{l}
14 - 5x \ge 0\\
9\left( {x - 2} \right)\left( {x + 6} \right) = {\left( {14 - 5x} \right)^2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{{14}}{5}\\
{x^2} - 11x + 19 = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{{14}}{5}\\
\left[ \begin{array}{l}
x = \dfrac{{11 + 3\sqrt 5 }}{2}\left( l \right)\\
x = \dfrac{{11 - 3\sqrt 5 }}{2}\left( c \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow x = \dfrac{{11 - 3\sqrt 5 }}{2}
\end{array}$
Vậy phương trình có tập nghiệm: $S = \left\{ {\dfrac{{11 - 3\sqrt 5 }}{2};3} \right\}$
b) ĐKXĐ: $x\ne\{0;1\}; x\ge \dfrac{-1}{3}$
Ta có:
$\begin{array}{l}
\dfrac{1}{{{{\left( {x - 1} \right)}^2}}} + \sqrt {3x + 1} = \dfrac{1}{{{x^2}}} + \sqrt {x + 2} \\
\Leftrightarrow \dfrac{1}{{{{\left( {x - 1} \right)}^2}}} - \dfrac{1}{{{x^2}}} + \sqrt {3x + 1} - \sqrt {x + 2} = 0\\
\Leftrightarrow \dfrac{{{x^2} - {{\left( {x - 1} \right)}^2}}}{{{x^2}{{\left( {x - 1} \right)}^2}}} + \dfrac{{3x + 1 - x - 2}}{{\sqrt {3x + 1} + \sqrt {x + 2} }} = 0\\
\Leftrightarrow \dfrac{{2x - 1}}{{{x^2}{{\left( {x - 1} \right)}^2}}} + \dfrac{{2x - 1}}{{\sqrt {3x + 1} + \sqrt {x + 2} }} = 0\\
\Leftrightarrow \left( {2x - 1} \right)\left( {\dfrac{1}{{{x^2}{{\left( {x - 1} \right)}^2}}} + \dfrac{1}{{\sqrt {3x + 1} + \sqrt {x + 2} }}} \right) = 0\\
\Leftrightarrow 2x - 1 = 0\left( {\dfrac{1}{{{x^2}{{\left( {x - 1} \right)}^2}}} + \dfrac{1}{{\sqrt {3x + 1} + \sqrt {x + 2} }} > 0,x \quad tm DKXD} \right)\\
\Leftrightarrow x = \dfrac{1}{2}\left( {tm} \right)
\end{array}$
Vậy phương trình có tập nghiệm là: $S = \left\{ {\dfrac{1}{2}} \right\}$
