Đáp án:
c) (x;y)=(-2;-3)
b) (x;y)=(3;-1)
Giải thích các bước giải:
\[\begin{array}{l}
c)\left\{ {\begin{array}{*{20}{c}}
{\frac{{3x}}{{x + 1}} - \frac{2}{{y + 4}} = 4}\\
{\frac{{2x}}{{x + 1}} + \frac{5}{{y + 4}} = 9}
\end{array}} \right.\\
dkxd:\left\{ {\begin{array}{*{20}{c}}
{x \ne - 1}\\
{y \ne - 4}
\end{array}} \right.\\
dat:\frac{x}{{x + 1}} = t;\frac{1}{{y + 4}} = v\\
hpt \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{3t - 2v = 4}\\
{2t + 5v = 9}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{t = 2}\\
{v = 1}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{\frac{x}{{x + 1}} = 2}\\
{\frac{1}{{y + 4}} = 1}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{2x + 2 = x}\\
{y + 4 = 1}
\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{x = - 2}\\
{y = - 3}
\end{array}} \right.} \right.} \right.} \right.} \right.\\
b)\left\{ {\begin{array}{*{20}{c}}
{x + y + \frac{4}{{x - y}} = 3}\\
{3x + 3y - \frac{8}{{x - y}} = 4}
\end{array}} \right.\\
dkxd:x \ne y\\
dat:x + y = t;\frac{1}{{x - y}} = v\\
hpt \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{t + 4v = 3}\\
{3t - 8v = 4}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{t = 2}\\
{v = \frac{1}{4}}
\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{x + y = 2}\\
{\frac{1}{{x - y}} = \frac{1}{4}}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x + y = 2}\\
{x - y = 4}
\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}
{x = 3}\\
{y = - 1}
\end{array}} \right.} \right.} \right.} \right.} \right.
\end{array}\]
