[tex]x-(2x-3) < 2(x-1,5)\\\\x-2x+3 < 2x-3\\\\-x+3 < 2x-3\\\\-x-2x < -3-3\\\\-3x < -6\\\\x > 2\\\\\boxed{x\in (2,+\infty )}[/tex]
Odpowiedź:
:)
[tex]x-(2x-3) < 2(x-1\frac{5}{10})\\\\x-2x+3 < 2(x-1\frac{1}{2})\\\\x-2x+3 < 2(x-\frac{3}{2})\\\\x-2x+3 < 2x-3\\\\x-2x-2x < -3-3\\\\-3x < -6\ \ \ /:(-3)\\\\x > 2[/tex]
x∈(2;+∞)