Solve the following equation using the method | vedclass

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(B) Given equation: $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{x}$Taking $LCM$ on the left side: $\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{5}{x

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$\{-\sqrt{5}, \sqrt{5}\}$

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(B) Given equation: $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{x}$Taking $LCM$ on the left side: $\frac{(x+1)^2 - (x-1)^2}{(x-1)(x+1)} = \frac{5}{x}$Simplify the numerator: $(x^2 + 2x + 1) - (x^2 - 2x + 1) = 4x$Simplify the denominator: $(x-1)(x+1) = x^2 - 1$So,$\frac{4x}{x^2 - 1} = \frac{5}{x}$Cross-multiply: $4x^2 = 5(x^2 - 1)$$4x^2 = 5x^2 - 5$$x^2 = 5$$x = \pm\sqrt{5}$Thus,the solution set is $\{-\sqrt{5}, \sqrt{5}\}$.

Solve the following equation using the method of factorization and write its solution set: $\frac{x+1}{x-1} - \frac{x-1}{x+1} = \frac{5}{x}$

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