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We are given the equation:

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

### Step 1: Factor the denominator on the left side
Notice that \( x^2 - 1 \) is a difference of squares, which can be factored as:
\[
x^2 - 1 = (x - 1)(x + 1)
\]
Thus, the equation becomes:
\[
\frac{2}{(x - 1)(x + 1)} + \frac{1}{1 - x} = 0
\]

### Step 2: Simplify the second term
Since \( 1 - x = -(x - 1) \), we rewrite the second term as:
\[
\frac{1}{1 - x} = \frac{-1}{x - 1}
\]
Now the equation becomes:
\[
\frac{2}{(x - 1)(x + 1)} - \frac{1}{x - 1} = 0
\]

### Step 3: Combine the terms
To combine the two fractions, we need a common denominator. The common denominator is \( (x - 1)(x + 1) \). Rewrite the second term with this denominator:
\[
\frac{-1}{x - 1} = \frac{-1(x + 1)}{(x - 1)(x + 1)} = \frac{-(x + 1)}{(x - 1)(x + 1)}
\]
Now the equation becomes:
\[
\frac{2}{(x - 1)(x + 1)} - \frac{x + 1}{(x - 1)(x + 1)} = 0
\]

### Step 4: Combine the numerators
Since the denominators are the same, we combine the numerators:
\[
\frac{2 - (x + 1)}{(x - 1)(x + 1)} = 0
\]
Simplifying the numerator:
\[
2 - (x + 1) = 2 - x - 1 = 1 - x
\]
Thus, the equation becomes:
\[
\frac{1 - x}{(x - 1)(x + 1)} = 0
\]

### Step 5: Solve the equation
For this fraction to equal zero, the numerator must be zero:
\[
1 - x = 0
\]
Solving for \( x \):
\[
x = 1
\]

### Step 6: Check for extraneous solutions
We should check if \( x = 1 \) makes any of the denominators zero. In the original equation, the denominators are \( x^2 - 1 = (x - 1)(x + 1) \) and \( 1 - x \). If \( x = 1 \), both denominators \( (x - 1) \) and \( 1 - x \) become zero, which leads to division by zero. Therefore, \( x = 1 \) is not a valid solution.

### Conclusion
Since \( x = 1 \) leads to division by zero, there are no valid solutions to this equation.

Would you like any further details or have questions on this process?

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Here are five related questions you might explore:

1. How do you solve rational equations with more than two terms?
2. What are extraneous solutions, and how do they occur?
3. How do you factor differences of squares in polynomial expressions?
4. Can you always simplify the expression \( \frac{1}{1 - x} = \frac{-1}{x - 1} \)?
5. How do you solve equations involving rational expressions that have higher powers of \( x \)?

**Tip:** Always check for extraneous solutions in rational equations by substituting the solutions back into the original denominators.

Solve for x. Show all steps to receive credit for your answer. (2/(x^2 - 1)) + (1/(1 - x)) = 0

Learn how to solve the rational equation (2/(x^2 - 1)) + (1/(1 - x)) = 0 with a detailed step-by-step solution. This problem covers key algebraic concepts, including factoring, simplifying rational expressions, and dealing with extraneous solutions. Suitable for students in Grades 10-12.

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