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(A) Given equation: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}$Taking the common denominator on the left side:$\frac{(x+1)^{2}-(x-1)^{2}}{(x-1)(x+1)

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$5$ and $-\frac{1}{5}$

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(A) Given equation: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6}$Taking the common denominator on the left side:$\frac{(x+1)^{2}-(x-1)^{2}}{(x-1)(x+1)}=\frac{5}{6}$Expanding the squares and simplifying the denominator:$\frac{(x^{2}+2x+1)-(x^{2}-2x+1)}{x^{2}-1}=\frac{5}{6}$$\frac{x^{2}+2x+1-x^{2}+2x-1}{x^{2}-1}=\frac{5}{6}$$\frac{4x}{x^{2}-1}=\frac{5}{6}$Cross-multiplying:$24x = 5(x^{2}-1)$$24x = 5x^{2}-5$Rearranging into standard quadratic form $ax^{2}+bx+c=0$:$5x^{2}-24x-5=0$Factorizing the quadratic equation:$5x^{2}-25x+x-5=0$$5x(x-5)+1(x-5)=0$$(x-5)(5x+1)=0$Setting each factor to zero:$x-5=0 \implies x=5$$5x+1=0 \implies x=-\frac{1}{5}$Thus,the roots are $5$ and $-\frac{1}{5}$.

Solve the following equation using the method of factorization: $\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{5}{6} \,(x \neq 1, -1)$

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