An algebra problem by A Former Brilliant Member

$\large \dfrac{x}{x^2-1}-\dfrac{1}{x^2-1}+\dfrac{2}{x+1}=0$

The $LCD$ is $x^2-1$ . So

$\large \dfrac{x-1+2(x-1)}{x^2-1}=0$ $\implies$ $\large x-1+2x-2=0$ $\implies$ $\large 3x=3$ $\implies$ $\large x=1$

Since $x=1$ makes $x^2-1=0$ in the given denominators, $x=1$ cannot be accepted as a root because division by $zero$ is not admissible. Hence, $1$ is an extraneous root and the given equation has $no~root$ .

Note that $\frac{x}{x^2 - 1} -\frac{1}{x^2 - 1}+\frac{2}{x+1} = \frac{x-1}{x^2 - 1}+\frac{2}{x+1}=\frac{x-1}{(x-1)(x+1)}+\frac{2}{x+1}=\frac{1}{x+1}+\frac{2}{x+1}=\frac{3}{x+1}$ which cannot be equal to $0$ for all real numbers $x$ . Therefore, the equation has no root .

I have another way.

$0=\frac { x }{ { x }^{ 2 }-1 } -\frac { 1 }{ { x }^{ 2 }-1 } +\frac { 2 }{ x+1 } \\ 0=\frac { x-1 }{ { x }^{ 2 }-1 } +\frac { 2 }{ x+1 } \\ 0=\frac { x-1 }{ (x+1)(x-1) } +\frac { 2 }{ x+1 } \\ 0=\frac { x-1+2(x-1) }{ (x+1)(x-1) } \\ 0=\frac { x-1+2x-2 }{ (x+1)(x-1) } \\ 0=\frac { 3x-3 }{ (x+1)(x-1) } \\ 0=\frac { 3(x-1) }{ (x+1)(x-1) } \\ 0=\frac { 3 }{ x+1 } \cdot { (x+1) }^{ 2 }\\ 0=3(x+1)\\ 0=x+1\\ -1=x\\$

Can anyone please tell me how can I make my own questions and post them on brilliant??