Đáp án:

\(\begin{array}{l}
1,\,\,\,\,C = \dfrac{3}{{\sqrt x  + 3}}\\
2,\,\,\,\,0\\
3,\\
a,\,\,\,x = 30\\
b,\,\,\,\,x = \dfrac{4}{3}
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
C = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{x - 9}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{{{\sqrt x }^2} - {3^2}}}\\
 = \dfrac{{\sqrt x }}{{\sqrt x  + 3}} + \dfrac{{2\sqrt x }}{{\sqrt x  - 3}} - \dfrac{{3x + 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 3} \right) + 2\sqrt x \left( {\sqrt x  + 3} \right) - \left( {3x + 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{\left( {x - 3\sqrt x } \right) + \left( {2x + 6\sqrt x } \right) - \left( {3x + 9} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{x - 3\sqrt x  + 2x + 6\sqrt x  - 3x - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3\sqrt x  - 9}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{{3.\left( {\sqrt x  - 3} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x  + 3} \right)}}\\
 = \dfrac{3}{{\sqrt x  + 3}}\\
2,\\
\left( {\sqrt {\dfrac{2}{3}}  + \sqrt {\dfrac{{50}}{3}}  - \sqrt {24} } \right).\sqrt 6 \\
 = \left( {\sqrt {\dfrac{1}{9}.6}  + \sqrt {\dfrac{{25}}{9}.6}  - \sqrt {4.6} } \right).\sqrt 6 \\
 = \left( {\sqrt {{{\left( {\dfrac{1}{3}} \right)}^2}.6}  + \sqrt {{{\left( {\dfrac{5}{3}} \right)}^2}.6}  - \sqrt {{2^2}.6} } \right).\sqrt 6 \\
 = \left( {\dfrac{1}{3}\sqrt 6  + \dfrac{5}{3}\sqrt 6  - 2\sqrt 6 } \right).\sqrt 6 \\
 = \sqrt 6 .\left( {\dfrac{1}{3} + \dfrac{5}{3} - 2} \right).\sqrt 6 \\
 = \sqrt 6 .0.\sqrt 6 \\
 = 0\\
3,\\
a,\\
DKXD:\,\,\,x \ge 5\\
\sqrt {4x - 20}  + \sqrt {x - 5}  - \dfrac{1}{3}\sqrt {9x - 45}  = 10\\
 \Leftrightarrow \sqrt {4.\left( {x - 5} \right)}  + \sqrt {x - 5}  - \dfrac{1}{3}\sqrt {9\left( {x - 5} \right)}  = 10\\
 \Leftrightarrow \sqrt {{2^2}.\left( {x - 5} \right)}  + \sqrt {x - 5}  - \dfrac{1}{3}.\sqrt {{3^2}.\left( {x - 5} \right)}  = 10\\
 \Leftrightarrow 2\sqrt {x - 5}  + \sqrt {x - 5}  - \dfrac{1}{3}.3\sqrt {x - 5}  = 10\\
 \Leftrightarrow 2\sqrt {x - 5}  + \sqrt {x - 5}  - \sqrt {x - 5}  = 10\\
 \Leftrightarrow 2\sqrt {x - 5}  = 10\\
 \Leftrightarrow \sqrt {x - 5}  = 5\\
 \Leftrightarrow x - 5 = {5^2}\\
 \Leftrightarrow x - 5 = 25\\
 \Leftrightarrow x = 30\\
b,\\
DKXD:\,\,\,{x^2} - 2x + 1 \ge 0 \Leftrightarrow {\left( {x - 1} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {{x^2} - 2x + 1}  = 3 - 2x\\
 \Leftrightarrow \sqrt {{x^2} - 2.x.1 + {1^2}}  = 3 - 2x\\
 \Leftrightarrow \sqrt {{{\left( {x - 1} \right)}^2}}  = 3 - 2x\\
 \Leftrightarrow \left| {x - 1} \right| = 3 - 2x\\
 \Leftrightarrow \left\{ \begin{array}{l}
3 - 2x \ge 0\\
\left[ \begin{array}{l}
x - 1 = 3 - 2x\\
x - 1 =  - \left( {3 - 2x} \right)
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
2x \le 3\\
x - 1 - 3 + 2x = 0\\
x - 1 + 3 - 2x = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{3}{2}\\
\left[ \begin{array}{l}
3x - 4 = 0\\
 - x + 2 = 0
\end{array} \right.
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \le \dfrac{3}{2}\\
\left[ \begin{array}{l}
x = \dfrac{4}{3}\\
x = 2
\end{array} \right.
\end{array} \right. \Leftrightarrow x = \dfrac{4}{3}
\end{array}\)