a, xy-2x-3y=5
⇒x(y-2)-3(y-2)=11
⇒(y-2)(x-3)=11
TH1: $\left \{ {{y-2=1} \atop {x-3=11}} \right.$ =>$\left \{ {{y=3} \atop {x=14}} \right.$
TH2: $\left \{ {{y-2=-1} \atop {x-3=-11}} \right.$ =>$\left \{ {{y=1} \atop {x=-8}} \right.$
TH3: $\left \{ {{y-2=11} \atop {x-3=1}} \right.$ =>$\left \{ {{y=13} \atop {x=4}} \right.$
TH4: $\left \{ {{y-2=-11} \atop {x-3=-1}} \right.$ =>$\left \{ {{y=-9} \atop {x=2}} \right.$
Vậy (x,y)∈{14;3);(-8;1);(4;13);(2;-9)}
b, Ta có: x²-5$\vdots$x+2
⇒x(x+2)-2(x+2)-1$\vdots$x+2
⇒x+2∈Ư(1)={±1}
x+2=1⇒x=-1
x+2=-1⇒x=-3
Vậy x∈{-1;-3}
c, Ta có: x²-5x-1$\vdots$x+3
⇒x(x+3)-8(x+3)+23$\vdots$x+3
⇒x+3∈Ư(23)={±1;±23}
x+3=1⇒x=-1
x+3=-1⇒x=-4
x+3=23⇒x=20
x+3=-23⇒x=-26
Vậy x∈{-1;-4;20;-26}
