Đáp án:

e) \(\left\{ \begin{array}{l}
x = \dfrac{{1 + \sqrt 5 }}{{1 + \sqrt 2 }}\\
y = \dfrac{{\sqrt {10}  - 1}}{{\sqrt {10}  + 1}}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a)\left\{ \begin{array}{l}
y = 7 - 2x\\
2x - 3.\left( {7 - 2x} \right) = 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 7 - 2x\\
8x = 22
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{11}}{4}\\
y = \dfrac{3}{2}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
x = 2 - 3y\\
 - 3\left( {2 - 3y} \right) - y = 5
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = 2 - 3y\\
8y = 11
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{{11}}{8}\\
x =  - \dfrac{{17}}{8}
\end{array} \right.\\
c)\left\{ \begin{array}{l}
y = 2x + 8\\
3x + 5\left( {2x + 8} \right) = 1
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = 2x + 8\\
13x =  - 39
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x =  - 3\\
y = 2
\end{array} \right.\\
d)DK:x \ne 1;y \ne 1\\
\left\{ \begin{array}{l}
\dfrac{{ - 5}}{{x - 1}} + \dfrac{1}{{y - 1}} = 10\\
\dfrac{1}{{x - 1}} + \dfrac{3}{{y - 1}} =  - 18
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{ - 5}}{{x - 1}} + \dfrac{1}{{y - 1}} = 10\\
\dfrac{5}{{x - 1}} + \dfrac{{15}}{{y - 1}} =  - 90
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\dfrac{{16}}{{y - 1}} =  - 80\\
\dfrac{{ - 5}}{{x - 1}} + \dfrac{1}{{y - 1}} = 10
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
y = \dfrac{4}{5}\\
x = \dfrac{2}{3}
\end{array} \right.\\
e)\left\{ \begin{array}{l}
\sqrt 2 x - \sqrt 5 y = 1\\
x + \sqrt 5 y = \sqrt 5 
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
\left( {1 + \sqrt 2 } \right)x = 1 + \sqrt 5 \\
x + \sqrt 5 y = \sqrt 5 
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{1 + \sqrt 5 }}{{1 + \sqrt 2 }}\\
\sqrt 5 y = \sqrt 5  - \dfrac{{1 + \sqrt 5 }}{{1 + \sqrt 2 }}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{1 + \sqrt 5 }}{{1 + \sqrt 2 }}\\
\sqrt 5 y = \dfrac{{\sqrt 5  + \sqrt {10}  - 1 - \sqrt 5 }}{{1 + \sqrt 2 }}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{1 + \sqrt 5 }}{{1 + \sqrt 2 }}\\
\sqrt 5 y = \dfrac{{\sqrt {10}  - 1}}{{1 + \sqrt 2 }}
\end{array} \right.\\
 \to \left\{ \begin{array}{l}
x = \dfrac{{1 + \sqrt 5 }}{{1 + \sqrt 2 }}\\
y = \dfrac{{\sqrt {10}  - 1}}{{\sqrt {10}  + 1}}
\end{array} \right.
\end{array}\)