a, ( x-1 )(y + 2 )= 5
TH1: $\left \{ {{x-1=1} \atop {y+2=5}} \right.$ =>$\left \{ {{x=2} \atop {y=3}} \right.$
TH2: $\left \{ {{x-1=-1} \atop {y+2=-5}} \right.$ =>$\left \{ {{x=0} \atop {y=-7}} \right.$
TH3: $\left \{ {{x-1=5} \atop {y+2=1}} \right.$ =>$\left \{ {{x=6} \atop {y=-1}} \right.$
TH4: $\left \{ {{x-1=-5} \atop {y+2=-1}} \right.$ =>$\left \{ {{x=-4} \atop {y=-3}} \right.$
Vậy (x,y)∈{(2;3);(0;-7);(6;-1);(-4;-3)}
b, ,( 2x-3 ) (y -1) = 12
TH1: $\left \{ {{2x-3=1} \atop {y-1=12}} \right.$ =>$\left \{ {{x=2} \atop {y=13}} \right.$
TH2: $\left \{ {{2x-3=-1} \atop {y-1=-12}} \right.$ =>$\left \{ {{x=1} \atop {y=-11}} \right.$
TH3: $\left \{ {{2x-3=12} \atop {y-1=1}} \right.$ =>$\left \{ {{x=15/2} \atop {y=2}} \right.$ (loại)
TH4: $\left \{ {{2x-3=-12} \atop {y-1=-1}} \right.$ =>$\left \{ {{x=-9/2} \atop {y=0}} \right.$ (loại)
TH5: $\left \{ {{2x-3=2} \atop {y-1=6}} \right.$ =>$\left \{ {{x=5/2} \atop {y=7}} \right.$ (loại)
TH6: $\left \{ {{2x-3=-2} \atop {y-1=-6}} \right.$ =>$\left \{ {{x=1/2} \atop {y=-5}} \right.$ (loại)
TH7: $\left \{ {{2x-3=6} \atop {y-1=2}} \right.$ =>$\left \{ {{x=3/2} \atop {y=3}} \right.$ (loại)
TH8: $\left \{ {{2x-3=-6} \atop {y-1=-2}} \right.$ =>$\left \{ {{x=-3/2} \atop {y=-1}} \right.$ (loại)
TH9: $\left \{ {{2x-3=4} \atop {y-1=3}} \right.$ =>$\left \{ {{x=7/2} \atop {y=4}} \right.$ (loại)
TH10: $\left \{ {{2x-3=-4} \atop {y-1=-3}} \right.$ =>$\left \{ {{x=-1/2} \atop {y=-2}} \right.$ (loại)
TH11: $\left \{ {{2x-3=3} \atop {y-1=4}} \right.$ =>$\left \{ {{x=3} \atop {y=5}} \right.$
TH12: $\left \{ {{2x-3=-3} \atop {y-1=-4}} \right.$ =>$\left \{ {{x=0} \atop {y=-3}} \right.$
Vậy (x,y)∈{(2;13);(1;-11);(3;5);(0;-3)}
c,( x-4 )( 2y+3 ) = 24
TH1: $\left \{ {{x-4=1} \atop {2y+3=24}} \right.$ =>$\left \{ {{x=5} \atop {y=21/2}} \right.$ (loại)
TH2: $\left \{ {{x-4=-1} \atop {2y+3=-24}} \right.$ =>$\left \{ {{x=3} \atop {y=-27/2}} \right.$ (loại)
TH3: $\left \{ {{x-4=24} \atop {2y+3=1}} \right.$ =>$\left \{ {{x=28} \atop {y=-1}} \right.$
TH4: $\left \{ {{x-4=-24} \atop {2y+3=-1}} \right.$ =>$\left \{ {{x=-20} \atop {y=-2}} \right.$
TH5: $\left \{ {{x-4=2} \atop {2y+3=12}} \right.$ =>$\left \{ {{x=6} \atop {y=9/2}} \right.$ (loại)
TH6: $\left \{ {{x-4=-2} \atop {2y+3=-12}} \right.$ =>$\left \{ {{x=2} \atop {y=-15/2}} \right.$ (loại)
TH7: $\left \{ {{x-4=12} \atop {2y+3=2}} \right.$ =>$\left \{ {{x=16} \atop {y=-1/2}} \right.$ (loại)
TH8: $\left \{ {{x-4=-12} \atop {2y+3=-2}} \right.$ =>$\left \{ {{x=-8} \atop {y=-5/2}} \right.$ (loại)
TH9: $\left \{ {{x-4=3} \atop {2y+3=8}} \right.$ =>$\left \{ {{x=7} \atop {y=5/2}} \right.$ (loại)
TH10: $\left \{ {{x-4=-3} \atop {2y+3=-8}} \right.$ =>$\left \{ {{x=1} \atop {y=-11/2}} \right.$ (loại)
TH11: $\left \{ {{x-4=8} \atop {2y+3=3}} \right.$ =>$\left \{ {{x=12} \atop {y=0}} \right.$
TH12: $\left \{ {{x-4=-8} \atop {2y+3=-3}} \right.$ =>$\left \{ {{x=-4} \atop {y=-3}} \right.$
TH13: $\left \{ {{x-4=4} \atop {2y+3=6}} \right.$ =>$\left \{ {{x=8} \atop {y=3/2}} \right.$ (loại)
TH14: $\left \{ {{x-4=-4} \atop {2y+3=-6}} \right.$ =>$\left \{ {{x=0} \atop {y=-9/2}} \right.$ (loại)
TH15: $\left \{ {{x-4=6} \atop {2y+3=4}} \right.$ =>$\left \{ {{x=10} \atop {y=1/2}} \right.$ (loại)
TH16: $\left \{ {{x-4=-6} \atop {2y+3=-4}} \right.$ =>$\left \{ {{x=-2} \atop {y=-7/2}} \right.$ (loại)
Vậy (x,y)∈{(28;-1);(-20;-2);(12;0);(-4;-3)}
d, xy+2x+y+4=0
⇒x(y+2)+y+4=0
⇒x(y+2)=-y-4
⇒x=$\frac{-y-4}{y+2}$
-Nếu: y+2=0⇒0x=-2 (loại)
-CHia 2 vế cho y+2∦0
⇒x=$\frac{-(y+2)-2}{y+2}$
⇒y+2∈Ư(2)={±1;±2}
y+2=1⇒y=-1⇒x=-3 (tm)
y+2=-1⇒y=-3⇒x=1 (tm)
y+2=2⇒y=0⇒x=-2 (tm)
y+2=-2⇒y=-4⇒x=0 (tm)
Vậy (x,y)∈{(-3;-1);(1;-3);(-2;0);(0;-4)}
e, xy+3x+y-3=0
⇒x(y+3)=3-y
-Nếu y+3=0⇒y=-3⇒0x=6 (loại)
-Chia 2 vế cho y+3∦0, ta có:
x=$\frac{3-y}{y+3}$
⇒x=$\frac{-(y+3)+3}{y+3}$
⇒y+3∈Ư(3)={±1;±3}
y+3=1⇒y=-2⇒x=2 (tm)
y+3=-1⇒y=-4⇒x=-4 (tm)
y+3=3⇒y=0⇒x=0 (tm)
y+3=-3⇒y=-6⇒x=-2 (tm)
Vậy (x,y)∈{(2;-2);(-4;-4);(0;0);(-2;-6)}
