Question

5/2+x<1+1/x-21/x+2-3/x-3<0

Answer 1

!!!!!!!!!!!!!!!!!!!!!!!!!!11

Answer 2

To solve the compound inequality $\frac{5}{2} + x < 1 + \frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3} < 0$, you'll want to break it down into two separate inequalities and solve each one individually.

Let's start with the leftmost inequality:

1. $\frac{5}{2} + x < 1 + \frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3}$

First, subtract 1 from both sides:

$\frac{5}{2} + x - 1 < \frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3}$

Simplify:

$\frac{3}{2} + x < \frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3}$

Now, let's work on the rightmost inequality:

2. $1 + \frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3} < 0$

First, subtract 1 from both sides:

$\frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3} - 1 < 0$

Simplify:

$\frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3} - 1 < 0$

Now you have two inequalities:

1. $\frac{3}{2} + x < \frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3}$

2. $\frac{1}{x} - \frac{21}{x+2} - \frac{3}{x-3} - 1 < 0$

You can solve each of these inequalities separately. Keep in mind that the solutions may be expressed as intervals or sets of values.