a) Ta có: \(x\left(y-7\right)+y-7=41\)
\(\Leftrightarrow\left(y-7\right)\left(x+1\right)=41\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y-7=41\\x+1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-7=-41\\x+1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y-7=1\\x+1=41\end{matrix}\right.\\\left\{{}\begin{matrix}y-7=-1\\x+1=-41\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=48\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=-34\\x=-2\end{matrix}\right.\\\left\{{}\begin{matrix}y=8\\x=40\end{matrix}\right.\\\left\{{}\begin{matrix}y=6\\x=-42\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(0;48);(-2;-34);(40;8);(-42;6)}
b) Ta có: \(x\left(y-3\right)-y+3=29\)
\(\Leftrightarrow x\left(y-3\right)-\left(y-3\right)=29\)
\(\Leftrightarrow\left(y-3\right)\left(x-1\right)=29\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y-3=29\\x-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y-3=-29\\x-1=-1\end{matrix}\right.\\\left\{{}\begin{matrix}y-3=1\\x-1=29\end{matrix}\right.\\\left\{{}\begin{matrix}y-3=-1\\x-1=-29\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=32\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-26\\x=0\end{matrix}\right.\\\left\{{}\begin{matrix}y=4\\x=30\end{matrix}\right.\\\left\{{}\begin{matrix}y=2\\x=-28\end{matrix}\right.\end{matrix}\right.\)
Vậy: (x,y)={(2;32);(0;-26);(30;4);(-28;2)}
a, x( y -7) + y - 7 = 41
x( y - 7) + ( y - 7) = 41
( y - 7)( x + 1) = 41
⇒ Cặp ( x ; y) ∈ ( 40 ; 2) , ( 0 ; 42) , ( -2 ; -40) , ( -42 ; 0)
