2 câu trả lời
Đáp án:
(y+2)x+1= y^2
=> xy+2x+1=y ²
=> x ²/4+2x+16-15=y ²-xy+x ²/4
=> (x/2+4) ²-15=(y-x/2) ²
=> (x-y+4)(y+4)=15
x,y nguyên => x-y+4, y+4 nguyên
=> y+4 ∈{-15,-5,-3,-1,1,3,5,15}
=> y ∈{-19, - 9,-7,-5,-3,-1,1,11}
y=-19=> x= -24
y=-9=> x=-16
y=-7=> x=-16
y=-5=> x=-10
y=-3=> x=8
y=-1=> x=0
y=1=>x=0
y=11=>x= 8
(y+2)x+1=y²
⇔(y+2)x-y²+1=0
⇔(y+2)x-y²+4=3
⇔(y+2)x-(y+2)(y-2)=3
⇔(y+2)(x-y+2)=3
⇒(y+2);(x-y+2)∈Ư(3)={±1;±3}
Lập bảng:
x -8 0 -8 0
x-y+2 -1 1 -3 3
y+2 -3 3 -1 1
y -5 1 -3 -1
Vậy (x;y)=(-8;-5),(0;1),(-2;-3),(2;-1).
