Question

1)(2x-5)^2-0,5x<(2x-1)(2x+1)-15 2)(12-1)(3x+1)<1+(6x+2)^2

Answer 1

решение : полм гл тмвволтд

Answer 2

To solve the inequalities, we'll work through each one step by step:

1. $(2x-5)^2 - 0.5x < (2x-1)(2x+1) - 15$

First, expand both sides of the inequality:

$4x^2 - 20x + 25 - 0.5x < 4x^2 - 1 - 15$

Simplify the terms:

$4x^2 - 20.5x + 25 < 4x^2 - 16$

Subtract $4x^2$ from both sides:

$-20.5x + 25 < -16$

Subtract 25 from both sides:

$-20.5x < -41$

Now, divide both sides by -20.5 (remember to reverse the inequality because we're dividing by a negative number):

$x > \frac{41}{20.5}$

Simplify:

$x > 2$

So, the solution for the first inequality is $x > 2$.

2. $(12-1)(3x+1) < 1 + (6x+2)^2$

Simplify the left side:

$11(3x+1) < 1 + (6x+2)^2$

Distribute 11:

$33x + 11 < 1 + (6x+2)^2$

Expand $(6x+2)^2$:

$33x + 11 < 1 + 36x^2 + 24x + 4$

Combine like terms:

$33x + 11 < 36x^2 + 24x + 5$

Subtract $33x$ and 11 from both sides:

$0 < 36x^2 - 9x - 6$

Divide the entire inequality by 3 to simplify:

$0 < 12x^2 - 3x - 2$

Now, we have a quadratic inequality. We can solve it by factoring or using the quadratic formula. Factoring might not work easily in this case, so let's use the quadratic formula:

The quadratic formula states that for an equation $ax^2 + bx + c = 0$, the solutions are given by:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In our case, $a = 12$, $b = -3$, and $c = -2$. Plugging these values in:

$x = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 12 \cdot (-2)}}{2 \cdot 12}$

Simplify the discriminant:

$x = \frac{3 \pm \sqrt{9 + 96}}{24}$

$x = \frac{3 \pm \sqrt{105}}{24}$

Since we're looking for values of $x$ that make the inequality $0 < 12x^2 - 3x - 2$ true, we need to find where this quadratic is positive. This means we'll consider the values of $x$ that satisfy:

$x > \frac{3 - \sqrt{105}}{24}$ and $x < \frac{3 + \sqrt{105}}{24}$

These are the two intervals of $x$ that satisfy the inequality.

So, the solution for the second inequality is:

$\frac{3 - \sqrt{105}}{24} < x < \frac{3 + \sqrt{105}}{24}$