To solve the equation 1/x-1 + 1/x+1 = 2/x^2-1, we multiply each term by (x-1)(x+1)(x^2-1) to clear the denominators:
(x-1)(x+1)(x^2-1)(1/x-1) + (x-1)(x+1)(x^2-1)(1/x+1) = (x-1)(x+1)(x^2-1)(2/x^2-1)
After distributing and canceling out common factors, the equation becomes:
(x+1)(x^2-1) + (x-1)(x^2-1) = 2(x-1)(x+1)
Simplifying this further, we get:
(x^3 + x^2 - x - 1) + (x^3 - x^2 - x + 1) = 2(x^2 - 1)
Combining like terms, we have:
2x^3 - 2x^2 - 2 = 2x^2 - 2
Bringing all terms to one side, we get:
2x^3 - 4x^2 = 0
Factoring out the common factor of 2x^2, we obtain:
2x^2(x - 2) = 0
Setting each factor equal to zero, we have two possible solutions:
1) 2x^2 = 0
x^2 = 0
x = 0
2) x - 2 = 0
x = 2
Therefore, the equation has two distinct solutions: x = 0 and x = 2.
Solve the following equation. If the equation has no solution or has infinitely many solutions, so state.
1/x-1 + 1/x+1 = 2/x^2-1
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