Đáp án:
d) \(\left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a\left\{ \begin{array}{l}
2x + 3y = - 6\\
- x + 5 = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 2\\
y = - \dfrac{{10}}{3}
\end{array} \right.\\
b)\left\{ \begin{array}{l}
- 3x + 5 = - 2y\\
2y + 4 = - 3x
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 3x + 2y = - 5\\
3x + 2y = - 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
4y = - 9\\
- 3x + 2y = - 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = - \dfrac{9}{4}\\
x = \dfrac{1}{6}
\end{array} \right.\\
c)\left\{ \begin{array}{l}
\sqrt 2 x + \sqrt 3 y = \sqrt 6 \\
x - y = \sqrt 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\sqrt 2 x + \sqrt 3 y = \sqrt 6 \\
\sqrt 3 x - \sqrt 3 y = \sqrt {15}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\left( {\sqrt 2 + \sqrt 3 } \right)x = \sqrt 6 + \sqrt {15} \\
x - y = \sqrt 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{\sqrt 6 + \sqrt {15} }}{{\sqrt 2 + \sqrt 3 }}\\
y = \dfrac{{\sqrt 6 + \sqrt {15} }}{{\sqrt 2 + \sqrt 3 }} - \sqrt 5 = \dfrac{{\sqrt 6 + \sqrt {15} - \sqrt {10} - \sqrt {15} }}{{\sqrt 2 + \sqrt 3 }}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{\sqrt 6 + \sqrt {15} }}{{\sqrt 2 + \sqrt 3 }}\\
y = \dfrac{{\sqrt 6 - \sqrt {10} }}{{\sqrt 2 + \sqrt 3 }}
\end{array} \right.\\
d)\left\{ \begin{array}{l}
x + y = 5\\
{x^2} - {y^2} = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + y = 5\\
\left( {x - y} \right)\left( {x + y} \right) = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + y = 5\\
5\left( {x - y} \right) = 5
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x + y = 5\\
x - y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x = 6\\
x - y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.
\end{array}\)
