Đáp án:
$\begin{array}{l}
i)Dkxd:\left\{ \begin{array}{l}
\\
x \ge 4
\end{array} \right.\\
\dfrac{{\sqrt {x + 4} + \sqrt {x - 4} }}{{\sqrt {x + 4} - \sqrt {x - 4} }} = 2\\
\Leftrightarrow \sqrt {x + 4} + \sqrt {x - 4} = 2\sqrt {x + 4} - 2\sqrt {x - 4} \\
\Leftrightarrow 3\sqrt {x - 4} = \sqrt {x + 4} \\
\Leftrightarrow 9\left( {x - 4} \right) = x + 4\\
\Leftrightarrow 9x - 36 = x + 4\\
\Leftrightarrow 9x - x = 4 + 36\\
\Leftrightarrow 8x = 40\\
\Leftrightarrow x = 5\left( {tmdk} \right)\\
Vậy\,x = 5\\
m)Dkxd:x \ge 0;x \ne 3\\
\dfrac{{\sqrt {{x^3} - 6{x^2} + 9x} }}{{x - 3}} = 2\\
\Leftrightarrow \sqrt {x\left( {{x^2} - 6x + 9} \right)} = 2\left( {x - 3} \right)\\
\Leftrightarrow \sqrt {x.{{\left( {x - 3} \right)}^2}} = 2\left( {x - 3} \right)\left( {dk:x > 3} \right)\\
\Leftrightarrow \sqrt x .\left( {x - 3} \right) = 2\left( {x - 3} \right)\\
\Leftrightarrow \sqrt x = 2\\
\Leftrightarrow x = 4\left( {tmdk} \right)\\
Vậy\,x = 4
\end{array}$
