\(\dfrac{1}{x-1}+1=\dfrac{x}{x-2}\\ \Rightarrow\dfrac{1+x-1}{x-1}=\dfrac{x}{x-2}\\ \dfrac{x}{x-1}=\dfrac{x}{x-2}\\ \Leftrightarrow x\left(x-2\right)=x\left(x-1\right)\\ x^2-2x=x^2-x\\ x^2-x^2=x-2x\\ -x=0\\ x=0\)
\(\Leftrightarrow x+2+x^2-3x+2=x^2-x\)
=>-2x+4+x=0
=>4-x=0
hay x=4