Đáp án:

f. \(\left\{ \begin{array}{l}
x = 9\\
y = 11
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
b.\left\{ \begin{array}{l}
2\sqrt 3 x + 3y = 3\\
\sqrt 2 x - 3y = \sqrt 2 
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
2\sqrt 3 x + \sqrt 2 x = 3 + \sqrt 2 \\
\sqrt 2 x - 3y = \sqrt 2 
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{3 + \sqrt 2 }}{{2\sqrt 3  + \sqrt 2 }}\\
y = \left( {\sqrt 2 .\dfrac{{3 + \sqrt 2 }}{{2\sqrt 3  + \sqrt 2 }} - \sqrt 2 } \right):3
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{3 + \sqrt 2 }}{{2\sqrt 3  + \sqrt 2 }}\\
y = \left( {\dfrac{{3\sqrt 2  + 2 - 2\sqrt 6  - 2}}{{2\sqrt 3  + \sqrt 2 }}} \right):3
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{3 + \sqrt 2 }}{{2\sqrt 3  + \sqrt 2 }}\\
y = \dfrac{{3\sqrt 2  - 2\sqrt 6 }}{{6\sqrt 3  + 3\sqrt 2 }}
\end{array} \right.\\
d.\left\{ \begin{array}{l}
y = \sqrt 3 x - \sqrt 3 \\
x + x\sqrt {15}  - \sqrt {15}  = \sqrt 5 
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{\sqrt 5  + \sqrt {15} }}{{1 + \sqrt {15} }}\\
y = \dfrac{{\sqrt {15}  + \sqrt {45}  - \sqrt 3  - \sqrt {45} }}{{1 + \sqrt {15} }}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{\sqrt 5  + \sqrt {15} }}{{1 + \sqrt {15} }}\\
y = \dfrac{{\sqrt {15}  - \sqrt 3 }}{{1 + \sqrt {15} }}
\end{array} \right.\\
e.DK:x \ne \dfrac{1}{2};y \ne 3\\
\left\{ \begin{array}{l}
\dfrac{3}{{2x - 1}} - \dfrac{6}{{3 - y}} =  - 1\\
\dfrac{3}{{2x - 1}} - \dfrac{3}{{3 - y}} = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
 - \dfrac{3}{{3 - y}} - \left( { - \dfrac{6}{{3 - y}}} \right) = 0 - \left( { - 1} \right)\\
\dfrac{3}{{2x - 1}} - \dfrac{3}{{3 - y}} = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{3}{{3 - y}} = 1\\
\dfrac{3}{{2x - 1}} - \dfrac{3}{{3 - y}} = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
3 - y = 3\\
\dfrac{3}{{2x - 1}} - \dfrac{3}{{3 - y}} = 0
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 0\\
x = 2
\end{array} \right.\\
f.\left\{ \begin{array}{l}
y = 29 - 2x\\
13x - 15\left( {29 - 2x} \right) =  - 48
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
43x = 387\\
y = 29 - 2x
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 9\\
y = 11
\end{array} \right.
\end{array}\)