Đáp án:
$\begin{array}{l}
8)\\
f\left( {5 - 2\sqrt 3 } \right) = f\left( 2 \right)\\
\Leftrightarrow \sqrt {5 - 2\sqrt 3 - 1} + m.\left( {5 - 2\sqrt 3 } \right) + 2\\
= \sqrt {2 - 1} + m.2 + 2\\
\Leftrightarrow \sqrt {4 - 2\sqrt 3 } + m.\left( {5 - 2\sqrt 3 } \right) = 1 + 2m\\
\Leftrightarrow \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + m\left( {5 - 2\sqrt 3 } \right) - 2m - 1 = 0\\
\Leftrightarrow \sqrt 3 - 1 + \left( {5 - 2\sqrt 3 - 2} \right).m - 1 = 0\\
\Leftrightarrow \left( {3 - 2\sqrt 3 } \right).m = 2 - \sqrt 3 \\
\Leftrightarrow \sqrt 3 .\left( {\sqrt 3 - 2} \right).m = 2 - \sqrt 3 \\
\Leftrightarrow m = - \dfrac{1}{{\sqrt 3 }} = - \dfrac{{\sqrt 3 }}{3}\\
Vậy\,m = - \dfrac{{\sqrt 3 }}{3}\\
9)\\
a)Dkxd:2x + 5\# 0\\
\Leftrightarrow x\# - \dfrac{5}{2}\\
Vậy\,x\# - \dfrac{5}{2}\\
b)Dkxd:\left\{ \begin{array}{l}
x + 1 \ge 0\\
2x - 3\# 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge - 1\\
x\# \dfrac{3}{2}
\end{array} \right.\\
Vậy\,x \ge - 1;x\# \dfrac{3}{2}\\
c)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
\sqrt x \# \dfrac{1}{2}
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x\# \dfrac{1}{4}
\end{array} \right.\\
Vậy\,x \ge 0;x\# \dfrac{1}{4}\\
d)Dkxd:\left\{ \begin{array}{l}
x \ge 0\\
5 - 2x\# 0\\
3 - x > 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \ge 0\\
x\# \dfrac{5}{2}\\
x < 3
\end{array} \right.\\
Vậy\,0 \le x < 3;x\# \dfrac{5}{2}
\end{array}$
