a.

$xy=-21=7.(-3)=(-7).3=3.(-7)=(-3).7=21.(-1)=(-21).1=(-1).21=1(-21)$

Do đó $(x,y)=(7,-3); (-7,3); (3,-7); (-3,7); (21,-1); (-21,1); (-1,21); (1,-21)$

b.

$(x+5)(y-3)=14=1.14=14.1=(-14)(-1)=(-1)(-14)=2.7=7.2=(-2)(-7)=(-7)(-2)$

Do đó:

$(x+5,y-3)=(1,14); (14,1); (-14,-1); (-1,-14); (2,7); (7,2); (-2,-7); (-7,-2)$

Đến đây thì đơn giản rồi.

c.

$x(y-2)=-19$, bạn làm tương tự

d. Tương tự

a)-3 x 7 =-21

-7 x 3=21

a) x.y=-21

x.y=-3.7=-7.3

x=-3,-7,3,7

y=7,3,-7,-3

b)(2x-1).(2y+1)=-35

-35=-5.7=-7.5

th1 (2x-1)=-5 suy ra x=(-5+1)/2=-2

      (2y+1)=7 suy ra x=(7-1)/2=3

th2 (2x-1)=7 suy ra x= 4

      (2y+1)=-5 suy ra x=-3

th3 (2x-1)=5 suy ra x = 3

      (2y+1)=-7 suy ra y = -4

th4 (2x-1)=-7 suy ra x= -3

      (2y+1)=5 suy ra x=2

Mình viết gọn thôi nhé , tại nhiều câu quá ^^

a/ \(\left(x+1\right)\left(1-y\right)=2\)

b/ \(\left(x+2\right)\left(y-1\right)=13\)

c/ \(\left(x-2\right)\left(y+3\right)=1\)

d/ \(\left(x-1\right)\left(y-1\right)=3\)

e/ \(\left(2x-y\right)\left(x+2y\right)=7\)

Về cách tìm nghiệm nguyên chắc bạn biết rồi nên mình không viết rõ ra nhé ^^

a: \(\left(x,y\right)\in\left\{\left(1;-21\right);\left(-21;1\right);\left(-1;21\right);\left(21;-1\right);\left(3;-7\right);\left(-7;3\right);\left(-3;7\right);\left(7;-3\right)\right\}\)

b: \(\Leftrightarrow\left(x,y-3\right)\in\left\{\left(1;-6\right);\left(-6;1\right);\left(2;-3\right);\left(-3;2\right);\left(-2;3\right);\left(3;-2\right);\left(6;-1\right);\left(-1;6\right)\right\}\)

hay \(\left(x,y\right)\in\left\{\left(1;-3\right);\left(-6;4\right);\left(2;0\right);\left(-3;-1\right);\left(-2;6\right);\left(3;1\right);\left(6;2\right);\left(-1;9\right)\right\}\)

a: (x-3)(2y+7)=1

=>(x-3;2y+7)∈{(1;1);(-1;-1)}

=>(x;2y)∈{(4;-6);(2;-8)}

=>(x;y)∈{(4;-3);(2;-4)}

b: (x+1)(y+2)=-3

=>(x+1;y+2)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(0;-5);(-4;-1);(-2;1);(2;-3)}

mà x<y

nên (x;y)∈{(-4;-1);(-2;1)}

c: xy+2x+y=-5

=>x(y+2)+y+2=-5+2

=>(x+1)(y+2)=-3

=>(x+1;y+2)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(0;-5);(-4;-1);(-2;1);(2;-3)}

a: xy+3x-7y=21

=>x(y+3)-7y-21=0

=>x(y+3)-7(y+3)=0

=>(x-7)(y+3)=0

=>\(\begin{cases}x-7=0\\ y+3=0\end{cases}\Rightarrow\begin{cases}x=7\\ y=-3\end{cases}\)

b: xy+3x-2y=11

=>x(y+3)-2y-6=11-6

=>x(y+3)-2(y+3)=5

=>(x-2)(y+3)=5

=>(x-2;y+3)∈{(1;5);(5;1);(-1;-5);(-5;-1)}

=>(x;y)∈{(3;2);(7;-2);(1;-8);(-3;-4)}