Given:-
[tex]\frac{4x-3}{x^2-2x}+\frac{-3x+3}{x^2-2x}[/tex]To find:-
The simplified form.
Since the denominators are same. we add the numerator simpily. so we get,
[tex]\begin{gathered} \frac{4x-3}{x^2-2x}+\frac{-3x+3}{x^2-2x}=\frac{4x-3-3x+3}{x^2-2x} \\ \text{ =}\frac{x}{x^2-2x} \end{gathered}[/tex]So now we cancel one x from the numerator and denominator. so we get,
[tex]\begin{gathered} \frac{x}{x^2-2x}=\frac{x}{x(x-2)} \\ \text{ =}\frac{1}{(x-2)} \end{gathered}[/tex]So the simplified value is,
[tex]\frac{1}{(x-2)}[/tex]