Đáp án:
$\begin{array}{l}
6)\sqrt {x - 1} .\left( {{x^2} - x - 6} \right) = 0\left( {dkxd:x \ge 1} \right)\\
\Leftrightarrow \sqrt {x - 1} .\left( {x - 3} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 3 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 3\\
x = - 2\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 1\,hoac\,x = 3\\
7)\\
\frac{{3{x^2} + 1}}{{\sqrt x - 1}} = \frac{4}{{x - 1}}\left( {dkxd:x > 1} \right)\\
\Rightarrow 3{x^2} + 1 = 4\\
\Rightarrow 3{x^2} = 3\\
\Rightarrow {x^2} = 1\\
\Rightarrow \left[ \begin{array}{l}
x = 1\left( {ktm} \right)\\
x = - 1\left( {ktm} \right)
\end{array} \right.\\
Vay\,x \in \emptyset \\
8)\frac{{{x^2} + 3x + 4}}{{\sqrt {x + 4} }} = \sqrt {x + 4} \left( {dkxd:x > - 4} \right)\\
\Rightarrow {x^2} + 3x + 4 = x + 4\\
\Rightarrow {x^2} + 2x = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = - 2
\end{array} \right.\\
Vay\,x = 0\,hoac\,x = - 2
\end{array}$
