Ta có : \(x+2y+xy=5\)
\(\Rightarrow x+y+xy=2.5\)
Từ đó ta có x= 5 và y =0
x +2y + xy = 5
=> x(y+1) + 2y + 2 = 5 + 2
=> x(y+1) + 2(y+1) = 7
=> (x+2)(y+1) = 7
=> bảng sau:
| x+2 | -1 | -7 | 1 | 7 |
| y+1 | -7 | -1 | 7 | 1 |
| x | -3 | -9 | -1 | 5 |
| y | -8 | -2 | 6 | 0 |
vậy các cặp (x;y) nguyên thỏa mãn là : (-3;-8);(-9;-2);(-1;6);(5;0)
(x+xy)+2y=5
x.(1+y)+2y-5=0
x(1+y)+2(y+1)-7=0
(1+y).(x+2)=7
Ta co:7= 1.7=-1.(-7)
+Neu 1+y=1thi y=0;x+2=7 thi x=5
+Neu 1+y=7 thi y=6; x+2=1thi x=-1(loai vi x la so nguyen am)
+Neu 1+y =-1 thi y=-2(loai vi y la so nguyen am)
+Neu 1+y=-7 thi y=-8(--------------------------------------)
Vay......
\(x+2y+xy=5\)
\(\Leftrightarrow x+\left(2y+xy\right)=5\)
\(\Rightarrow x+y\left(2+x\right)=5\)
\(\Rightarrow\left(2+x\right)+y\left(2+x\right)=5+2=7\)
\(\Rightarrow\left(2+x\right)+\left(1+y\right)=7\)
\(\Rightarrow\frac{2+x=1;7;-1;-7}{1+y=7;1;-7;-1}\)
\(\Rightarrow\frac{x=-1;5;-3;-9}{y=6;0;-8;-2}\)
\(Ta\)\(có\)\(:\)\(x+2y+xy=5\)
\(\Rightarrow\)\(x+\left(2y+xy\right)=5\)
\(\Rightarrow\)\(x+2+y\left(2+x\right)=5+2\)
\(\Rightarrow\)\(\left(x+2\right)\left(1+y\right)=7=1.7=\left(-1\right)\left(-7\right)\)
| \(x+2\) | \(1\) | \(7\) | \(-1\) | \(-7\) |
| \(1+y\) | \(7\) | \(1\) | \(-7\) | \(-1\) |
\(\Rightarrow\)
| \(x\) | \(-1\) | \(5\) | \(-3\) | \(-9\) |
| \(y\) | \(6\) | \(0\) | \(-8\) | \(-2\) |
\(Vì\)\(x,y\)\(là\)\(các\)\(số\)\(nguyên\)\(không\)\(âm\)\(nên\)\(x=5,y=0.\)
