To solve the equation \(\frac{2x}{x+1} + \frac{5}{2x} = 2\), we can follow these steps:

### Step 1: Eliminate the fractions
Multiply the entire equation by the least common denominator (LCD) to eliminate the fractions. The denominators here are \(x+1\) and \(2x\), so the LCD is \(2x(x+1)\).

Multiply each term by \(2x(x+1)\):

\[
2x(x+1) \cdot \frac{2x}{x+1} + 2x(x+1) \cdot \frac{5}{2x} = 2x(x+1) \cdot 2
\]

Simplifying each term:

\[
4x^2 + 5(x+1) = 4x(x+1)
\]

### Step 2: Expand and simplify
Expand both sides of the equation:

\[
4x^2 + 5x + 5 = 4x^2 + 4x
\]

### Step 3: Move all terms to one side
Subtract \(4x^2\) from both sides to simplify:

\[
5x + 5 = 4x
\]

### Step 4: Solve for \(x\)
Subtract \(4x\) from both sides:

\[
x + 5 = 0
\]

Now, solve for \(x\):

\[
x = -5
\]

### Step 5: Check for extraneous solutions
Substitute \(x = -5\) back into the original equation to ensure it is not an extraneous solution.

The original equation is:

\[
\frac{2x}{x+1} + \frac{5}{2x} = 2
\]

Substitute \(x = -5\):

\[
\frac{2(-5)}{-5+1} + \frac{5}{2(-5)} = \frac{-10}{-4} + \frac{5}{-10} = \frac{5}{2} - \frac{1}{2} = 2
\]

Since the original equation holds true, the solution is:

\[
\boxed{-5}
\]

Would you like more details on any part of this process, or do you have any questions?

### Related Questions:
1. How would the solution change if the original equation had a different constant term on the right-hand side?
2. Can you solve the equation using a substitution method?
3. What if the equation was \(\frac{2x}{x+1} + \frac{5}{2x} = 3\)? How would that affect the solution?
4. How can we verify the solution using a graphing approach?
5. What are the implications if \(x\) was constrained to be positive?

### Tip:
Always check for extraneous solutions when solving equations involving fractions, especially when multiplying by an expression that contains variables.

Discover how to solve the rational equation 2x/(x+1) + 5/(2x) = 2 with a comprehensive step-by-step guide. This problem involves key algebraic techniques such as eliminating fractions and checking for extraneous solutions. Ideal for students in Grades 9-11.

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