https://file.aiquickdraw.com/mathbot/file/1725468894931vhum8b7d.jpg

The inequality provided in the image is:

\[
\frac{x - 1}{2x + 5} > \frac{-3x}{5 - 2x} - \frac{-2x}{2x - 5}
\]

To solve this inequality, let's proceed step by step:

### Step 1: Simplify the right-hand side
We first need to combine the fractions on the right-hand side.

The right-hand side is:

\[
\frac{-3x}{5 - 2x} - \frac{-2x}{2x - 5}
\]

Notice that \(5 - 2x\) and \(2x - 5\) are negatives of each other:

\[
5 - 2x = -(2x - 5)
\]

Thus, we can rewrite the right-hand side as:

\[
\frac{-3x}{5 - 2x} - \frac{-2x}{-(5 - 2x)} = \frac{-3x}{5 - 2x} + \frac{2x}{5 - 2x}
\]

Combine the fractions:

\[
\frac{-3x + 2x}{5 - 2x} = \frac{-x}{5 - 2x}
\]

### Step 2: Rewrite the inequality
Now, the inequality becomes:

\[
\frac{x - 1}{2x + 5} > \frac{-x}{5 - 2x}
\]

### Step 3: Cross-multiply (taking into account sign changes)
To solve the inequality, we can cross-multiply. However, be careful: the sign of the inequality might change depending on the sign of the expressions \(2x + 5\) and \(5 - 2x\).

- \(2x + 5\) is positive when \(x > -\frac{5}{2}\).
- \(5 - 2x\) is positive when \(x < \frac{5}{2}\).

So, we need to consider the intervals \(x < -\frac{5}{2}\), \(-\frac{5}{2} < x < \frac{5}{2}\), and \(x > \frac{5}{2}\).

#### Case 1: \(x < -\frac{5}{2}\)
In this case, both \(2x + 5\) and \(5 - 2x\) are negative, so we multiply by their absolute values and flip the inequality:

\[
-(x - 1)(5 - 2x) < -x(2x + 5)
\]

Simplifying this, we get:

\[
(x - 1)(5 - 2x) > x(2x + 5)
\]

Expanding both sides:

\[
5x - 2x^2 - 5 + 2x > 2x^2 + 5x
\]

Combine like terms:

\[
-4x^2 + 7x - 5 > 0
\]

#### Case 2: \(-\frac{5}{2} < x < \frac{5}{2}\)
In this interval, \(2x + 5 > 0\) and \(5 - 2x > 0\), so we do not flip the inequality:

\[
(x - 1)(5 - 2x) > -x(2x + 5)
\]

Expanding:

\[
5x - 2x^2 - 5 + 2x > -2x^2 - 5x
\]

Combine like terms:

\[
-4x^2 + 7x - 5 > 0
\]

This is the same quadratic as above.

#### Case 3: \(x > \frac{5}{2}\)
Here, both \(2x + 5 > 0\) and \(5 - 2x < 0\), so we flip the inequality:

\[
(x - 1)(5 - 2x) < -x(2x + 5)
\]

Expanding:

\[
5x - 2x^2 - 5 + 2x < -2x^2 - 5x
\]

This simplifies to:

\[
-4x^2 + 7x - 5 < 0
\]

### Conclusion
We now need to solve the quadratic inequality \(-4x^2 + 7x - 5 > 0\) and \(-4x^2 + 7x - 5 < 0\) in the respective intervals.

Would you like to proceed with the solution for the quadratic inequality, or is there any specific step you'd like to delve into further?

Solve the inequality: (x - 1)/(2x + 5) > (-3x)/(5 - 2x) - (-2x)/(2x - 5)

This detailed solution covers the process of solving a complex algebraic inequality that involves combining fractions, cross-multiplying, and solving quadratic inequalities. Suitable for high school students in Grades 10-12.

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