a: xy-2x-3=20
=>xy-2x=23
=>x(y-2)=23
=>\(x\left(y-2\right)=1\cdot23=23\cdot1=\left(-1\right)\cdot\left(-23\right)=\left(-23\right)\cdot\left(-1\right)\)
=>\(\left(x;y-2\right)\in\left\{\left(1;23\right);\left(23;1\right);\left(-1;-23\right);\left(-23;-1\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(1;25\right);\left(23;3\right);\left(-1;-21\right);\left(-23;1\right)\right\}\)
b: 2xy+3y-15=5
=>2xy+3y=20
=>\(y\left(2x+3\right)=20\)
mà 2x+3 lẻ(do x nguyên)
nên \(\left(2x+3\right)\cdot y=1\cdot20=5\cdot4=\left(-1\right)\cdot\left(-20\right)=\left(-5\right)\cdot\left(-4\right)\)
=>\(\left(2x+3;y\right)\in\left\{\left(1;20\right);\left(5;4\right);\left(-1;-20\right);\left(-5;-4\right)\right\}\)
=>\(\left(2x;y\right)\in\left\{\left(-2;20\right);\left(2;4\right);\left(-4;-20\right);\left(-8;-4\right)\right\}\)
=>\(\left(x;y\right)\in\left\{\left(-1;20\right);\left(1;2\right);\left(-2;-20\right);\left(-4;-4\right)\right\}\)
a) x(y - 2) - 3 = 20
⇒ x(y - 2) = 23
Ta có bảng sau:
| y - 2 | 23 | 1 | -23 | -1 |
| x | 1 | 23 | -1 | -23 |
| y | 25 | 3 | -21 | 1 |
b) 2xy + 3y - 15 = 5
⇒ y(2x + 3) = 15 + 5
⇒ y(2x + 3) = 20
Mà x nguyên nên 2x + 3 là số lẻ ⇒ 2x + 3 ∈ {1; -1; 5; -5}
⇒ x ∈ {-1; -2; 1; -4}
Khi x = -1 ⇒ y = 20
Khi x = -2 ⇒ y = -20
Khi x = 1 ⇒ y = 4
Khi x = -4 ⇒ y = -4