Đáp án:
Giải thích các bước giải:
`a,xy+x+y=16`
`⇔x(y+1)+y-16=0`
`⇔x(y+1)+(y+1)-17=0`
`⇔(y+1)(x+1)=17`
+)TH1: $\left \{ {{y+1=1} \atop {x+1=17}} \right.$
$⇔\left \{ {{y=0} \atop {x=16}} \right.$
+)TH2: $\left \{ {{y+1=-1} \atop {x+1=-17}} \right.$
$⇔\left \{ {{y=-2} \atop {x=-18}} \right.$
Vậy `x=16;y=0` hoặc `x=-18;y=-2`
`b,xy+3x-7y=21`
`⇔(xy+3x)-7y-21=0`
`⇔(xy+3x)-(7y+21)=0`
`⇔x(y+3)-7(y+3)=0`
`⇔(y+3)(x-7)=0`
`⇔`\(\left[ \begin{array}{l}y+3=0\\x-7=0\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}y=-3\\x=7\end{array} \right.\)
Vậy `x=7;y=-3`
