Đáp án:a)(x;y)=(4;3); (2;-4); (10;0) (-4;-1)
b)(x;y)=(2;-3); (5;-1); (-1;19) (-28;1)
Giải thích các bước giải:
a)(x-3)(2y+1)=7
⇒ x-3; 2y+1∈Ư(7)
⇒x-3; 2y+1∈ Ư={-1;1;-7;7}
⇔(1)$\left \{ {{x-3=1} \atop {2y+1=7}} \right.$ ⇔$\left \{ {{x=4} \atop {y=3}} \right.$
(2)$\left \{ {{x-3=-1} \atop {2y+1=-7}} \right.$ ⇔$\left \{ {{x=2} \atop {y=-4}} \right.$
(3)$\left \{ {{x-3=7} \atop {2y+1=1}} \right.$ ⇔$\left \{ {{x=10} \atop {y=0}} \right.$
(4)$\left \{ {{x-3=-7} \atop {2y+1=-1}} \right.$ ⇔$\left \{ {{x=-4} \atop {y=-1}} \right.$
⇒ (x;y)=(4;3); (2;-4); (10;0) (-4;-1)
b)(2x+1).(3y-2)=-55
⇒ 2x+1; 3y-2∈Ư(-55)
⇒2x+1; 3y-2∈ Ư={-1;1;-5;5;-11;11;55;-55}
⇔(1)$\left \{ {{2x+1=-5} \atop {3y-2=11}} \right.$ ⇔$\left \{ {{x=-3} \atop {y=\frac{13}{3}}} \right.$ (loại)
⇔(2)$\left \{ {{2x+1=5} \atop {3y-2=-11}} \right.$ ⇔$\left \{ {{x=2} \atop {y=-3}} \right.$
⇔(3)$\left \{ {{2x+1=-11} \atop {3y-2=5}} \right.$ ⇔$\left \{ {{x=6} \atop {y=\frac{7}{3}}} \right.$ (loại)
⇔(4)$\left \{ {{2x+1=11} \atop {3y-2=-5}} \right.$ ⇔$\left \{ {{x=5} \atop {y=-1}} \right.$
⇔(5)$\left \{ {{2x+1=-1} \atop {3y-2=55}} \right.$ ⇔$\left \{ {{x=-1} \atop {y=19}} \right.$
⇔(6)$\left \{ {{2x+1=1} \atop {3y-2=-55}} \right.$ ⇔$\left \{ {{x=0} \atop {y=\frac{-53}{3}}} \right.$(loại)
⇔(7) $\left \{ {{2x+1=-55} \atop {3y-2=1}} \right.$ ⇔$\left \{ {{x=-28} \atop {y=1}} \right.$
⇔(8)$\left \{ {{2x+1=55} \atop {3y-2=-1}} \right.$ ⇔$\left \{ {{x=27} \atop {y=\frac{1}{3}}} \right.$ (loại)
⇒(x;y)=(2;-3); (5;-1); (-1;19) (-28;1)
