Question

Tìm \(x,y\in z\) thỏa:

a) \(xy+2x+2y=3\)

b) \(y\left(x+1\right)=3x+5\)

Answer 1

a)

\(xy+2x+2y=3\)

=> \(xy+2x+2y+4=7\)

=> \(x\left(y+2\right)+2\left(y+2\right)=7\)

=> \(\left(x+2\right)\left(y+2\right)=7\)

=> \(\left\{{}\begin{matrix}x+2=1\\y+2=7\end{matrix}\right.\)=> \(x=-1;y=5\)

=> \(\left\{{}\begin{matrix}x+2=7\\y+2=1\end{matrix}\right.\)=> \(x=5;y=-1\)

=> \(\left\{{}\begin{matrix}x+2=-1\\y+2=-7\end{matrix}\right.\)=> \(x=-3;y=-9\)

=> \(\left\{{}\begin{matrix}x+2=-7\\y+2=-1\end{matrix}\right.\)=> \(x=-9;y=-3\)

Answer 2

b)

\(y\left(x+1\right)=3x+5\)

=> \(xy+y=3x+5\)

=> \(xy+y-3x-5=0\)

=> \(x\left(y-3\right)+y-5=0\)

=> \(x\left(y-3\right)+y-3=2\)

=> \(\left(x+1\right)\left(y-3\right)=2\)

=> \(\left\{{}\begin{matrix}x+1=1\\y-3=2\end{matrix}\right.\)=> \(x=0;y=5\)

=> \(\left\{{}\begin{matrix}x+1=2\\y-3=1\end{matrix}\right.\)=> \(x=1;y=4\)

=> \(\left\{{}\begin{matrix}x+1=-1\\y-3=-2\end{matrix}\right.\)=> \(x=-2;y=1\)

=> \(\left\{{}\begin{matrix}x+1=-2\\y-3=-1\end{matrix}\right.\)=> \(x=-3;y=2\)

Answer 3

\(xy+2x+2y=3\)

\(\Rightarrow xy+2x+2y+4=7\)

\(\Rightarrow x\left(y+2\right)+2\left(y+2\right)=7\)

\(\Rightarrow\left(x+2\right)\left(y+2\right)=7\)

\(\Rightarrow x+2;y+2\in U\left(7\right)\)

\(U\left(7\right)=\left\{\pm1;\pm7\right\}\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+2=1\Rightarrow x=-1\\y+2=7\Rightarrow y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x+2=-1\Rightarrow x=-3\\y+2=-7\Rightarrow y=-9\end{matrix}\right.\\\left\{{}\begin{matrix}x+2=7\Rightarrow x=5\\y+2=1\Rightarrow y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x+2=-7\Rightarrow x=-9\\y+2=-1\Rightarrow y=-3\end{matrix}\right.\end{matrix}\right.\)

\(y\left(y+1\right)=3x+5\)

\(\Rightarrow xy+y=3x+5\)

\(\Rightarrow xy+y-3x=5\)

\(\Rightarrow xy+y-3x-3=2\)

\(\Rightarrow x\left(y-3\right)+1\left(y-3\right)=2\)

\(\Rightarrow\left(x+1\right)\left(y-3\right)=2\)

\(\Rightarrow x+1;y-3\in U\left(2\right)\)

\(U\left(2\right)=\left\{\pm1;\pm2\right\}\)

Xét ước ~~~