We are tasked with solving the inequality:

\[
\frac{x-2}{x+2} + \frac{1}{x-1} \geq 0
\]

### Step 1: Combine the two fractions.

First, we need to find a common denominator to combine the two fractions. The denominators are \( (x+2) \) and \( (x-1) \). The least common denominator (LCD) is their product: \( (x+2)(x-1) \).

Rewrite the expression with the common denominator:

\[
\frac{(x-2)}{(x+2)} + \frac{1}{(x-1)} = \frac{(x-2)(x-1) + (x+2)}{(x+2)(x-1)}
\]

### Step 2: Simplify the numerator.

Now, expand the terms in the numerator:

\[
(x-2)(x-1) = x^2 - x - 2x + 2 = x^2 - 3x + 2
\]

So, we now have:

\[
\frac{x^2 - 3x + 2 + x + 2}{(x+2)(x-1)} = \frac{x^2 - 2x + 4}{(x+2)(x-1)}
\]

### Step 3: Solve the inequality.

We now need to solve:

\[
\frac{x^2 - 2x + 4}{(x+2)(x-1)} \geq 0
\]

To solve this inequality, we need to examine the critical points where the numerator and denominator are zero, and determine where the expression is positive or negative.

#### Critical points:
- The denominator is zero when \( x = -2 \) or \( x = 1 \).
- The numerator \( x^2 - 2x + 4 \) is a quadratic with a discriminant \( \Delta = (-2)^2 - 4(1)(4) = 4 - 16 = -12 \). Since the discriminant is negative, the numerator has no real roots and is always positive.

Thus, the sign of the expression depends entirely on the denominator. The denominator changes signs at \( x = -2 \) and \( x = 1 \).

#### Sign Analysis:

We analyze the sign of the expression in the intervals determined by \( x = -2 \) and \( x = 1 \).

1. For \( x < -2 \), both factors of the denominator, \( (x+2) \) and \( (x-1) \), are negative. Thus, the entire expression is positive.
2. For \( -2 < x < 1 \), \( (x+2) \) is positive and \( (x-1) \) is negative. The expression is negative.
3. For \( x > 1 \), both factors of the denominator are positive, making the entire expression positive.

### Step 4: Conclusion.

The expression is non-negative (i.e., \( \geq 0 \)) in the intervals where the fraction is positive. The expression is positive for \( x < -2 \) or \( x > 1 \). However, we must exclude \( x = -2 \) and \( x = 1 \) because the denominator would be zero, causing the expression to be undefined at those points.

Thus, the solution is:

\[
x \in (-\infty, -2) \cup (1, \infty)
\]

Would you like further details on any part of the solution?

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Here are 5 related questions for practice:
1. Solve \( \frac{2x+3}{x-4} \geq 1 \).
2. Solve \( \frac{x+1}{x^2-1} > 0 \).
3. Solve \( \frac{2x-1}{x+2} \leq 0 \).
4. Solve \( \frac{3x+2}{x^2 - 9} \geq 0 \).
5. Solve \( \frac{x^2 + 3x - 4}{x+1} \geq 0 \).

**Tip**: Always check for points where the denominator is zero in rational inequalities.

Learn how to solve the rational inequality \frac{x-2}{x+2} + \frac{1}{x-1} \geq 0 with a step-by-step solution. This solution covers combining rational expressions, solving quadratic inequalities, and performing sign analysis to find the critical points. Ideal for students in Grades 10-12.

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