Đáp án: x=2
Giải thích các bước giải:
$\begin{array}{l}
\left( {x - 1} \right)\left( {x + 2} \right) + 2\left( {x - 1} \right).\sqrt {\frac{{x + 2}}{{x - 1}}} = 8\\
Đkxđ:x > 1\\
\Rightarrow \left( {x - 1} \right)\left( {x + 2} \right) + 2\sqrt {\left( {x - 1} \right)\left( {x + 2} \right)} = 8\\
\Rightarrow {\left( {\sqrt {\left( {x - 1} \right)\left( {x + 2} \right)} } \right)^2} + 2\sqrt {\left( {x + 1} \right)\left( {x + 2} \right)} - 8 = 0\\
\Rightarrow \left[ \begin{array}{l}
\sqrt {\left( {x - 1} \right)\left( {x + 2} \right)} = 2\\
\sqrt {\left( {x - 1} \right)\left( {x + 2} \right)} = - 4\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow \left( {x - 1} \right)\left( {x + 2} \right) = 4\\
\Rightarrow {x^2} + x - 6 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 3} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\left( {tm} \right)\\
x = - 3\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 2
\end{array}$
