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a: \(\left(x,y\right)\in\left\{\left(-9;1\right);\left(-1;9\right);\left(-3;3\right)\right\}\)
b: \(\left(x,y\right)\in\left\{\left(1;7\right);\left(-7;-1\right)\right\}\)
c: \(\left(x,y\right)\in\left\{\left(11;-1\right);\left(-1;11\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>xy-x+y=0
=>x(y-1)+y-1=-1
=>(y-1)(x+1)=-1
=>(x+1;y-1) thuộc {(1;-1); (-1;1)}
=>(x,y) thuộc {(0;0); (-2;2)}
b: =>x(y+2)+y-1=0
=>x(y+2)+y+2-3=0
=>(y+2)(x+1)=3
=>(x+1;y+2) thuộc {(1;3); (3;1); (-1;-3); (-3;-1)}
=>(x,y) thuộc {(0;1); (2;-1); (-2;-5); (-4;-3)}
c:
y>=3
=>y+5>=8
=>y(x-7)+5x-35=-35
=>(x-7)(y+5)=-35
mà y+5>=8
nên (y+5;x-7) thuộc (35;-1)
=>(y;x) thuộc {(30;6)}
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có ; -5x = -15 y/5 = 2
=> x = -15:(-5) => y:5 = 2
=> x = 3 (1) => y = 5x2
=> y = 10 (2)
Từ (1) và (2)
=> x + y = 3 + 10 = 13
=> x - y = 3 - 10 = (-7)
ok rồi nhé
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì -5x=-15 nên:
x=-15:(-5)
x=3
Vì \(\frac{y}{5}\)=2 nên y=2.5
y=10
\(\Rightarrow\)x+y=3+10=13
\(\Rightarrow\)x-y=3-10=-7
![](https://rs.olm.vn/images/avt/0.png?1311)
a: xy=x-y
=>xy-x+y=0
=>xy-x+y-1=-1
=>x(y-1)+(y-1)=-1
=>(x+1)(y-1)=-1
=>\(\left(x+1\right)\left(y-1\right)=1\cdot\left(-1\right)=\left(-1\right)\cdot1\)
=>\(\left(x+1;y-1\right)\in\left\{\left(1;-1\right);\left(-1;1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;0\right);\left(-2;2\right)\right\}\)
b: x(y+2)+y=1
=>\(x\left(y+2\right)+y+2=3\)
=>\(\left(x+1\right)\left(y+2\right)=3\)
=>\(\left(x+1\right)\cdot\left(y+2\right)=1\cdot3=3\cdot1=\left(-1\right)\left(-3\right)=\left(-3\right)\left(-1\right)\)
=>\(\left(x+1;y+2\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)
=>\(\left(x,y\right)\in\left\{\left(0;1\right);\left(2;-1\right);\left(-2;-5\right);\left(-4;-3\right)\right\}\)
a, x +y =-3+[-2]=5
b,x-y=-3-[-2]=-3+2=-1
Ta có : (-5).x=15
=>x=15:(-5)
=>x=(-3)
22.y=(-44)
=>y=(-44):22
=>y=(-2)
Ta lại có: x+y=(-3)+(-2)=(-5)
x-y=(-3)-(-2)=(-1)