a ) \(\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy -..
b ) \(\left(x-5\right)\left(y+1\right)=7\)
Sẽ có 4 trường hợp xảy ra :
TH1 :
\(\left[{}\begin{matrix}x-5=7\\y+1=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\y=0\end{matrix}\right.\)
TH2 :
\(\left[{}\begin{matrix}x-5=1\\y+1=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\y=6\end{matrix}\right.\)
TH3 :
\(\left[{}\begin{matrix}x-5=-7\\y+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\y=0\end{matrix}\right.\)
TH4 :
\(\left[{}\begin{matrix}x-5=-1\\y+1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\y=-8\end{matrix}\right.\)
Vậy -..
