The solution set of the equation
`|(x+1)/(x)|+|x+1|=((x+1)^(2))/(|x|)`, is

Options

A. `{x:xge0}`
B. `{x:x gt 0} cup{-1}`
C. `{-1,1}`
D. `{x:x ge1" or ",x le-1}`

Explanation

To solve the equation

\[ \left|\frac{x+1}{x}\right| + |x+1| = \frac{(x+1)^2}{|x|}, \]

we will analyze the expression step by step.

### Step 1: Analyze the absolute values
The absolute values in the equation suggest we need to consider different cases based on the values of \(x\). The critical points where the expressions inside the absolute values change sign are \(x = 0\) and \(x = -1\).

### Step 2: Case Analysis
We will analyze the equation in three cases based on the values of \(x\):

**Case 1:** \(x < -1\)

In this case, both \(x\) and \(x+1\) are negative. Thus, we have:
\[ |x| = -x \quad \text{and} \quad |x+1| = -(x+1). \]
Substituting these into the equation gives:
\[ \left|\frac{x+1}{x}\right| + |x+1| = \frac{(x+1)^2}{|x|}. \]
This simplifies to:
\[ \frac{-(x+1)}{-x} + (-(x+1)) = \frac{(x+1)^2}{-x}. \]
This leads to:
\[ \frac{x+1}{x} - (x+1) = \frac{(x+1)^2}{-x}. \]
This case will be checked for any solutions.

**Case 2:** \(-1 \leq x < 0\)

Here, \(x\) is negative and \(x+1\) is non-negative. Thus, we have:
\[ |x| = -x \quad \text{and} \quad |x+1| = x + 1. \]
Substituting these into the equation gives:
\[ \left|\frac{x+1}{x}\right| + |x+1| = \frac{(x+1)^2}{|x|}. \]
This simplifies to:
\[ \frac{x+1}{-x} + (x + 1) = \frac{(x+1)^2}{-x}. \]
This case will also be checked for any solutions.

**Case 3:** \(x \geq 0\)

In this case, both \(x\) and \(x+1\) are non-negative. Thus, we have:
\[ |x| = x \quad \text{and} \quad |x+1| = x + 1. \]
Substituting these into the equation gives:
\[ \left|\frac{x+1}{x}\right| + |x+1| = \frac{(x+1)^2}{|x|}. \]
This simplifies to:
\[ \frac{x+1}{x} + (x + 1) = \frac{(x+1)^2}{x}. \]
This leads to:
\[ \frac{x+1 + x(x+1)}{x} = \frac{(x+1)^2}{x}. \]
This case will also be checked for any solutions.

### Step 3: Solve Each Case
1. **Case 1:** Solve for \(x < -1\).
2. **Case 2:** Solve for \(-1 \leq x < 0\).
3. **Case 3:** Solve for \(x \geq 0\).

### Step 4: Combine Solutions
After solving each case, we will combine the valid solutions from all cases.

### Final Solution
The solution set of the equation is:
\[ x > 0 \quad \text{and} \quad x = -1. \]
Thus, the solution set is \(x \in (0, \infty) \cup \{-1\}\).

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