xy + x - y = 6
\(\Rightarrow x\left(y+1\right)-y-1=5\)
\(\Rightarrow x\left(y+1\right)-\left(y+1\right)=5\)
\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)
Ta có bảng sau:
| x - 1 | 1 | -1 | 5 | -5 |
| y + 1 | 5 | -5 | 1 | -1 |
| z | 2 | 0 | 6 | -4 |
| y | 4 | -6 | 0 | -2 |
Vậy...
=> x(y+1) - (y+1) = 5
=> (x-1)(y+1) = 5
sau tự giải nha bn , k mik nữa
Ta có : \(xy+x-y=6\)
\(\Leftrightarrow xy+x=6+y\)
\(\Leftrightarrow x\left(y+1\right)=6+y\)
\(\Rightarrow\left(x-1\right)\left(y+1\right)=5\)
\(\cdot th1:x-1=-1\Rightarrow x=0;y=-6\)
\(\cdot th2:x-1=1\Rightarrow x=2;y=4\)
\(\cdot th3:x-1=-5\Rightarrow x=-6;y=-2\)
\(\cdot th4:x=-1=5\Rightarrow x=6;y=0\)
x.(y+1)+y=6
x.(y+1)+y+1=6+1
x.(y+1)+(y+1)=7
(x+1).(y+1)=7 = 1.7=7.1=(-1).(-7)=(-7).(-1)
TH1 : (x+1).(y+1)=1.7
=>x=0 ; y=6
TH2 : (x+1).(y+1)=7.1
=> x=6 ; y=0
TH3: (x+1).(y+1)=(-1).(-7)
=> x=-2 ; y=-8
TH3: (x+1).(y+1)=(-7).(-1)
=> x=-8 ; y=-2
mk không chắc là nó đúng đâu !!!
