Giải thích các bước giải:

\(\begin{array}{l}
a)dk:\left\{ {\begin{array}{*{20}{c}}
{x \ne  - 1}\\
{y \ne 1}
\end{array}} \right.\\
\left\{ {\begin{array}{*{20}{c}}
{\frac{{6x - 3}}{{y - 1}} - \frac{{2y}}{{x + 1}} = 5}\\
{\frac{{4x - 2}}{{y - 1}} - \frac{{4y}}{{x + 1}} = 2}
\end{array} \Leftrightarrow } \right.\left\{ {\begin{array}{*{20}{c}}
{\frac{{3(2x - 1)}}{{y - 1}} - \frac{{2y}}{{x + 1}} = 5}\\
{\frac{{2(2x - 1)}}{{y - 1}} - \frac{{4y}}{{x + 1}} = 2}
\end{array}} \right.\\
dat:\frac{{2x - 1}}{{y - 1}} = a;\frac{y}{{x + 1}} = b\\
pt \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{3a - 2b = 5}\\
{2a - 4b = 2}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{a = 2}\\
{b = \frac{1}{2}}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{\frac{{2x - 1}}{{y - 1}} = 2}\\
{\frac{y}{{x + 1}} = \frac{1}{2}}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{2x - 2y =  - 1}\\
{x - 2y =  - 1}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x = 0}\\
{y = \frac{1}{2}}
\end{array}} \right.} \right.} \right.} \right.} \right.\\
c)\left\{ {\begin{array}{*{20}{c}}
{(x + 5)(y - 2) = xy}\\
{(x - 5)(y + 12) = xy}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{xy - 2x + 5y - 10 = xy}\\
{xy + 12x - 5y - 60 = xy}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{2x - 5y =  - 10}\\
{12x - 5y = 60}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x = 7}\\
{y = \frac{{24}}{5}}
\end{array}} \right.} \right.} \right.} \right.\\
f)\left\{ {\begin{array}{*{20}{c}}
{x + y = \frac{{4x - 3}}{5}}\\
{x + 3y = \frac{{15 - 9y}}{{14}}}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{5x + 5y = 4x - 3}\\
{14x + 42y = 15 - 9y}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x + 5y =  - 3}\\
{14x + 51y = 15}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x = 12}\\
{y =  - 3}
\end{array}} \right.} \right.} \right.\\
i)dk:\left\{ {\begin{array}{*{20}{c}}
{x > 0}\\
{y > 0}
\end{array}} \right.\\
\left\{ {\begin{array}{*{20}{c}}
{\frac{4}{{\sqrt x }} + \frac{3}{{\sqrt y }} = \frac{{13}}{{36}}}\\
{\frac{6}{{\sqrt x }} + \frac{{10}}{{\sqrt y }} = 1}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{{\sqrt x }} = \frac{1}{{36}}}\\
{\frac{1}{{\sqrt y }} = \frac{1}{{12}}}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{\sqrt x  = 36}\\
{\sqrt y  = 12}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x = 1296}\\
{y = 144}
\end{array}} \right.
\end{array}\)