Giải thích các bước giải:
a, (5x-1)(y-4)=4
⇒ 5x-1;y-4 ∈Ư(4)={±1;±2;±4}
ta thấy : 5x-1 chia 5 dư 4
⇒ 5x-1 ∈{-1;4}
⇒ y-4 ∈{-4;1}
TH1 : \(\left[ \begin{array}{l}5x-1=-1⇒5x=0⇒x=0 (TM)\\y-4=-4⇒y=0 (TM)\end{array} \right.\)
TH2 : \(\left[ \begin{array}{l}5x-1=4⇒5x=5⇒x=1 (TM)\\y-4=1⇒y=5 (TM)\end{array} \right.\)
Vậy (x;y)=(0;0);(1;5)
xy+x+2y=5
⇒ x(y+1)+2y=5
⇒ x(y+1)+2.y+2.1=7
⇒ x(y+1)+2(y+1)=7
⇒ (x+2)(y+1)=7
⇒ x+2;y+1 ∈Ư(7)={±1;±7}
ta có
x+2 1 7 -1 -7
y+1 7 1 -7 -1
x -1 5 -3 -9
y 6 0 -8 -2
Vậy (x;y)=(-1;6);(5;0);(-3;-8);(-9;-2)
