Đáp án:
h) \(\left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{4}{x} - \dfrac{4}{y} = 4\\
\dfrac{3}{x} + \dfrac{4}{y} = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{7}{x} = 9\\
\dfrac{4}{x} - \dfrac{4}{y} = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{7}{9}\\
y = \dfrac{7}{2}
\end{array} \right.\\
d)DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{3}{x} + \dfrac{3}{y} = \dfrac{1}{8}\\
\dfrac{2}{x} - \dfrac{3}{y} = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{x} = \dfrac{1}{8}\\
\dfrac{2}{x} - \dfrac{3}{y} = 0
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 40\\
y = 60
\end{array} \right.\\
g)DK:x \ne 0;y \ne - 12\\
\left\{ \begin{array}{l}
\dfrac{{40}}{x} - \dfrac{5}{{y + 12}} = 5\\
\dfrac{1}{x} + \dfrac{5}{{y + 12}} = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{41}}{x} = 8\\
\dfrac{1}{x} + \dfrac{5}{{y + 12}} = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{41}}{8}\\
y = - \dfrac{{235}}{{23}}
\end{array} \right.\\
b)DK:x \ne 0;y \ne 0\\
\left\{ \begin{array}{l}
\dfrac{{12}}{x} + \dfrac{{10}}{y} = 6\\
\dfrac{9}{x} - \dfrac{{10}}{y} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{21}}{x} = 7\\
\dfrac{9}{x} - \dfrac{{10}}{y} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 3\\
y = 5
\end{array} \right.\\
e)DK:x \ne 2;y \ne 1\\
\left\{ \begin{array}{l}
\dfrac{3}{{x - 2}} + \dfrac{3}{{y - 1}} = 6\\
\dfrac{2}{{x - 2}} - \dfrac{3}{{y - 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{{x - 2}} = 7\\
\dfrac{2}{{x - 2}} - \dfrac{3}{{y - 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{19}}{7}\\
y = \dfrac{8}{3}
\end{array} \right.\\
h)DK:x \ne - \dfrac{1}{2};y \ne 1\\
\left\{ \begin{array}{l}
\dfrac{8}{{2x + 1}} + \dfrac{{18}}{{y - 1}} = - 2\\
\dfrac{{27}}{{2x + 1}} - \dfrac{{18}}{{y - 1}} = \dfrac{{39}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{35}}{{2x + 1}} = \dfrac{{35}}{2}\\
\dfrac{{27}}{{2x + 1}} - \dfrac{{18}}{{y - 1}} = \dfrac{{39}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x + 1 = 2\\
\dfrac{{27}}{{2x + 1}} - \dfrac{{18}}{{y - 1}} = \dfrac{{39}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{2}\\
y = - 2
\end{array} \right.
\end{array}\)
( Những câu còn lại bạn dùng cộng đại số và làm tương tự nhé )
