a) (2x - y)(x + 2y) = -3
1) 2x - y = 3 => y = 2x - 3
x + 2y = -1 => x + 2(2x - 3) = - 1 =>
x + 4x - 6 = -1 => 5x = 5 => x = 1
y = 2x - 3 = 2 - 3 = -1.
Первое решение (x, y) = (1, -1)
2) 2x - y = - 3 => y = 2x + 3
x + 2y = 1 => x +2(2x + 3) = 1 =>
x + 4x + 6 = 1 => 5x = - 5 => x = -1
y = 2x + 3 = -2 + 3 = 1.
Второе решение (x, y) = (-1, 1)
3) 2x - y = 1 => y = 2x - 1
x + 2y = -3 => x + 2(2x - 1) = -3 =>
x + 4x - 2 = -3 => 5x = -1 => x = - 1/5 - нецелое.
4) 2x - y = -1 => y = 2x + 1
x + 2y = 3 => x + 2(2x + 1) = 3 =>
x + 4x + 2 = 3 => 5x = 1 => x = 1/5 - нецелое.
Всего два решения (x, y) = (1, -1) и (x, y) = (-1, 1)
б) 2x² + xy - y² = -3
x² - y² + x² + xy = -3
(x - y)(x + y) + x(x + y) = -3
(x + y)(x - y + x) = -3
(x + y)(2x - y) = -3
1) x + y = 3 => y = 3 - x
2x - y = -1 => 2x - 3 + x = -1 =>
3x - 3 = -1 => 3x = 2 => x = 2/3 - нецелое
2) x + y = 1 => y = 1 - x
2x - y = -3 => 2x - 1 + x = -3 =>
3x -1 = -3 => 3x = -2 => x = -2/3 - нецелое
3) x + y = -3 => y = -3 - x
2x - y = 1 => 2x +3 + x = 1 =>
3x + 3 = 1 = 3x = -2 => x = -2/3 - нецелое
4) x + y = -1 = y = -1 - x
2x - y = 3 => 2x +1 + x = 3 =>
3x + 1 = 3 => 3x = 2 => x = 2/3 - нецелое.
Второе уравнение не имеет решений в целых числах.
(x-1)(x+2)<0
x^2+x-2<0 корни -2 и 1 по т. Виета и получаем x^2+x-2=(x-1)(x+2)
неравенства равносильны.