Question

(x + 1)(x-4) >(x-2)(x-5)​

Answer 1

$(x + 1)(x-4) >(x-2)(x-5)\\x^2-4x+x-4>x^2-5x-2x+10\\x^2-3x-4>x^2-7x+10\\-3x-4>-7x+10\\7x-3x>4+10\\4x>14\\x>\frac{14}{4} \\x>3.5$

Ответ: $x\in(3.5;+\infty)$

Answer 2

To solve the inequality $(x + 1)(x - 4) > (x - 2)(x - 5)$, we'll follow these steps:

1. Expand both sides of the inequality.
2. Simplify the inequality.
3. Find the critical points (where the expression is equal to zero) and test the intervals between these points to determine the solution.

Step 1: Expand both sides of the inequality. $(x + 1)(x - 4) = x^2 - 3x - 4$ $(x - 2)(x - 5) = x^2 - 7x + 10$

Step 2: Rewrite the inequality using the expanded expressions. $x^2 - 3x - 4 > x^2 - 7x + 10$

Step 3: Simplify the inequality. To solve the inequality, we can get rid of the common $x^2$ term on both sides:

$x^2 - x^2 - 3x + 7x > 10 + 4$

$4x > 14$

Finally, divide both sides by 4:

$x > \frac{14}{4}$

$x > \frac{7}{2}$

So the solution to the inequality is $x > \frac{7}{2}$. In interval notation, this is $(\frac{7}{2}, +\infty)$.