Solution-
The equation given here,
[tex]\frac{1}{x-1} =\frac{x-2}{2x^2-2}[/tex]
[tex]\Rightarrow 2x^2-2=(x-1)(x-2)[/tex]
[tex]\Rightarrow 2x^2-2=x^2-3x+2[/tex]
[tex]\Rightarrow x^2+3x-4=0[/tex]
[tex]\Rightarrow x^2 +4x-x-4=0[/tex]
[tex]\Rightarrow (x-1)(x+4)=0[/tex]
[tex]\Rightarrow x=1 \ and -4[/tex]
But, at x=1, [tex]\frac{1}{x-1} \ and \ \frac{x-2}{2x^2-2}[/tex] becomes infinity so it can't be a solution to the equation.
∴ x=1 is the extraneous solution of the equation
