Giải thích các bước giải:
\(\begin{array}{l}
a)\, \Leftrightarrow 3y\left( {2x + 3} \right) + 10x + 15 = 17\\
\Leftrightarrow 3y\left( {2x + 3} \right) + 5\left( {2x + 3} \right) = 17\\
\Leftrightarrow \left( {2x + 3} \right)\left( {3y + 5} \right) = 17\\
TH1:\,\left\{ \begin{array}{l}
2x + 3 = 1\\
3y + 5 = 17
\end{array} \right. \Rightarrow \left\{ \begin{array}{l}
x = - 1\\
y = 4
\end{array} \right.\left( L \right)\\
TH2:\,\left\{ \begin{array}{l}
2x + 3 = 17\\
3y + 5 = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 7\\
y = - \frac{4}{3}
\end{array} \right.\left( L \right)\\
c)\, \Leftrightarrow 9xy - 6x - 15y = 21\\
\Leftrightarrow 3x\left( {3y - 2} \right) - 5\left( {3y - 2} \right) = 31\\
\Leftrightarrow \left( {3x - 5} \right)\left( {3y - 2} \right) = 31\\
TH1:\,\left\{ \begin{array}{l}
3x - 5 = 1\\
3y - 2 = 31
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 11
\end{array} \right.\left( {tm} \right)\\
TH2:\,\left\{ \begin{array}{l}
3x - 5 = - 1\\
3y - 2 = - 31
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = \frac{4}{3}\\
y = \frac{{ - 29}}{3}
\end{array} \right.\left( {ktm} \right)\\
TH3:\,\left\{ \begin{array}{l}
3x - 5 = 31\\
3y - 2 = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 12\\
y = 1
\end{array} \right.\left( {tm} \right)\\
TH4:\,\left\{ \begin{array}{l}
3x - 5 = - 31\\
3y - 2 = - 1
\end{array} \right.\left( {ktm} \right)
\end{array}\)
Tương tự với câu b em nhé
