a/ (2x-1)(y+2)=0
⇔ \(\left[ \begin{array}{l}2x-1=0\\y+2=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=\frac{1}{2}\\y=-2\end{array} \right.\)
b/ (x-3)(xy+2)=7
TH1:
\(\left[ \begin{array}{l}x-3=1\\xy+2=7\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\xy=5\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\4y=5\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=4\\y=\frac{5}{4}\end{array} \right.\)
TH2:
\(\left[ \begin{array}{l}x-3=7\\xy+2=1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=10\\xy=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=10\\10y=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=10\\y=\frac{-1}{10}\end{array} \right.\)
TH3:
\(\left[ \begin{array}{l}x-3=-1\\xy+2=-7\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\xy=-9\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\2y=-9\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=2\\y=\frac{-9}{2}\end{array} \right.\)
TH4:
\(\left[ \begin{array}{l}x-3=-7\\xy+2=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-4\\xy=-3\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-4\\-4y=-1\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=-4\\y=\frac{1}{4}\end{array} \right.\)
Xin hay nhất ạ
