Question

{(X+5)(x-2)-(x-4)(x+4)<12 7x-1>_—2

Answer 1

Ответ:

x>5/126

Объяснение:

{(X+5)(x-2)-(x-4)(x+4)<12 7x-1>_—2

Answer 2

It seems like there are a couple of inequalities and expressions in your input. Let's break it down step by step.

1. Simplify the Expressions:

   Let's start by simplifying the expressions:

   a. $(X+5)(x-2)-(x-4)(x+4)$:

   Using the distributive property, we can expand both terms:

   $(X+5)(x-2) = X(x) + X(-2) + 5(x) + 5(-2) = X^2 + 3X - 10$

   $(x-4)(x+4) = x(x) + x(4) - 4(x) - 4(4) = x^2 - 4x - 16$

   Subtracting the second expression from the first:

   $(X+5)(x-2)-(x-4)(x+4) = (X^2 + 3X - 10) - (x^2 - 4x - 16) = X^2 + 3X - 10 - x^2 + 4x + 16 = -x^2 + 7x + 6$

   b. $7x - 1$:

   This expression is already simplified.

2. Solving the Inequalities:

   a. $-x^2 + 7x + 6 < 12$:

   Subtracting 12 from both sides:

   $-x^2 + 7x + 6 - 12 < 0$

   $-x^2 + 7x - 6 < 0$

   This is a quadratic inequality. To solve it, you can find the critical points by setting the quadratic expression equal to zero:

   $-x^2 + 7x - 6 = 0$

   Factoring, you get:

   $-(x - 6)(x + 1) = 0$

   So, the critical points are $x = 6$ and $x = -1$.

   Now, test the intervals determined by these critical points with test points to determine the sign of the quadratic expression in those intervals. You'll find that the inequality is satisfied when $-1 < x < 6$.

   b. $7x - 1 > -2$:

   Adding 1 to both sides:

   $7x - 1 + 1 > -2 + 1$

   $7x > -1$

   Dividing both sides by 7 (since 7 is positive):

   $x > -\frac{1}{7}$

So, the solutions to the inequalities are:

a. $-1 < x < 6$ b. $x > -\frac{1}{7}$

Please double-check the calculations and steps, as mathematical accuracy is important.