[tex]\dfrac x{x-2}+\dfrac{x-1}{x+1}=1[/tex]
[tex]\dfrac x{x-2}\cdot\dfrac{x+1}{x+1}+\dfrac{x-1}{x+1}\cdot\dfrac{x-2}{x-2}=1[/tex]
So long as [tex]x\neq-1[/tex] and [tex]x\neq2[/tex], we can carry out the manipulation above. Then
[tex]\dfrac{x(x+1)}{(x-2)(x+1)}+\dfrac{(x-1)(x-2)}{(x+1)(x-2)}=1[/tex]
[tex]\dfrac{x(x+1)+(x-1)(x-2)}{(x-2)(x+1)}=1[/tex]
[tex]\dfrac{x^2+x+x^2-3x+2}{x^2-x-2}=1[/tex]
[tex]\dfrac{2x^2-2x+2}{x^2-x-2}=1[/tex]
[tex]\dfrac{2x^2-2x+2}{x^2-x-2}\cdot(x^2-x-2)=1\cdot(x^2-x-2)[/tex]
[tex]2x^2-2x+2=x^2-x-2[/tex]
[tex]x^2-x+4=0[/tex]
We can complete the square to solve:
[tex]x^2-x+\dfrac14+\dfrac{15}4=0[/tex]
[tex]\left(x-\dfrac12\right)^2=-\dfrac{15}4[/tex]
However, [tex]y^2\ge0[/tex] for all (real) values of [tex]y[/tex], which means there is no real solution to this equation.
If you're solving over the complex numbers, we can take the square root of both sides to get
[tex]x-\dfrac12=\pm i\dfrac{\sqrt{15}}2[/tex]
[tex]\implies x=\dfrac{1\pm i\sqrt{15}}2[/tex]
