Đáp án:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
y = - \dfrac{1}{5}\\
x = - \dfrac{1}{5}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
y = - 1\\
x = 3
\end{array} \right.\\
c)\left\{ \begin{array}{l}
x = \dfrac{{13}}{4}\\
y = 6
\end{array} \right.\\
d)\left\{ \begin{array}{l}
x = - \dfrac{9}{4}\\
y = - \dfrac{7}{8}
\end{array} \right.
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left\{ \begin{array}{l}
2x - 7y = 1\\
21x - 6y = - 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
42x - 147y = 21\\
42x - 12y = - 6
\end{array} \right.\\
\to \left\{ \begin{array}{l}
135y = - 27\\
x = \dfrac{{1 + 7y}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{1}{5}\\
x = - \dfrac{1}{5}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
2x + 3y = 3\\
7x + 5y = 16
\end{array} \right.\\
\to \left\{ \begin{array}{l}
14x + 21y = 21\\
14x + 10y = 32
\end{array} \right.\\
\to \left\{ \begin{array}{l}
11y = - 11\\
x = \dfrac{{3 - 3y}}{2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - 1\\
x = 3
\end{array} \right.\\
c)DK:x \ne 2;y \ne 1\\
\left\{ \begin{array}{l}
\dfrac{1}{{x - 2}} + \dfrac{1}{{y - 1}} = 1\\
\dfrac{2}{{x - 2}} - \dfrac{3}{{y - 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{3}{{x - 2}} + \dfrac{3}{{y - 1}} = 3\\
\dfrac{2}{{x - 2}} - \dfrac{3}{{y - 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{5}{{x - 2}} = 4\\
\dfrac{1}{{x - 2}} + \dfrac{1}{{y - 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{1}{{x - 2}} = \dfrac{4}{5}\\
\dfrac{4}{5} + \dfrac{1}{{y - 1}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x - 2 = \dfrac{5}{4}\\
y - 1 = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{13}}{4}\\
y = 6
\end{array} \right.\\
d)DK:x \ne \pm 2y\\
\left\{ \begin{array}{l}
\dfrac{4}{{x + 2y}} - \dfrac{1}{{x - 2y}} = 1\\
\dfrac{{20}}{{x + 2y}} + \dfrac{3}{{x - 2y}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{{20}}{{x + 2y}} - \dfrac{5}{{x - 2y}} = 5\\
\dfrac{{20}}{{x + 2y}} + \dfrac{3}{{x - 2y}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{8}{{x - 2y}} = - 4\\
\dfrac{4}{{x + 2y}} - \dfrac{1}{{x - 2y}} = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{1}{{x - 2y}} = - 2\\
\dfrac{4}{{x + 2y}} - \left( { - 2} \right) = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x - 2y = - \dfrac{1}{2}\\
x + 2y = - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x = - \dfrac{9}{2}\\
x + 2y = - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - \dfrac{9}{4}\\
y = - \dfrac{7}{8}
\end{array} \right.
\end{array}\)
