Đáp án:
5) \(\left\{ \begin{array}{l}
x = 1\\
y = 0
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
3)DK:x \ne - 1;y \ne - 4\\
\left\{ \begin{array}{l}
\dfrac{{15x}}{{x + 1}} - \dfrac{{10}}{{y + 4}} = 20\\
\dfrac{{4x}}{{x + 1}} - \dfrac{{10}}{{y + 4}} = 18
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\dfrac{x}{{x + 1}} = 2\\
\dfrac{{4x}}{{x + 1}} - \dfrac{{10}}{{y + 4}} = 18
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - 2\\
4.2 - \dfrac{{10}}{{y + 4}} = 18
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - 2\\
y + 4 = - 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = - 2\\
y = - 5
\end{array} \right.\\
4)DK:x \ne \dfrac{3}{2};y \ne \dfrac{5}{2}\\
\left\{ \begin{array}{l}
2x - 3y = - 16\\
{\left( {2x - 3} \right)^2} = {\left( {2y - 5} \right)^2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x + 16 = 3y\\
{\left( {2x - 3} \right)^2} = {\left( {2y - 5} \right)^2}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x - 3 = 3y - 19\\
{\left( {3y - 19} \right)^2} = 4{y^2} - 20y + 25
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x - 3 = 3y - 19\\
9{y^2} - 114y + 361 = 4{y^2} - 20y + 25
\end{array} \right.\\
\to \left\{ \begin{array}{l}
2x - 3 = 3y - 19\\
\left[ \begin{array}{l}
y = 14\\
y = \dfrac{{24}}{5}
\end{array} \right.
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 13\\
x = - \dfrac{4}{5}
\end{array} \right.\\
5)DK:x \ne 0\\
\left\{ \begin{array}{l}
1 + 2x\left( {x + y} \right) = 3x\\
3x\left( {x + y} \right) - x = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
6x\left( {x + y} \right) - 9x = - 3\\
6x\left( {x + y} \right) - 2x = 4
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 7x = - 7\\
3x\left( {x + y} \right) - x = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = 1\\
y = 0
\end{array} \right.
\end{array}\)
