Question 356193
{{{1/(x^2-x)=1/x}}}
You need to use a common denominator, {{{x^2-x=x(x-1)}}}
{{{1/(x^2-x)=(x-1)/(x(x-1))}}}
{{{1/(x^2-x)-(x-1)/(x(x-1))=0}}}
{{{(1-(x-1))/(x(x-1))=0}}}
{{{(1-x+1)/(x(x-1))=0}}}
{{{(2-x)/(x(x-1))=0}}}
The expression equals zero when,
{{{2-x=0}}}
{{{highlight(x=2)}}}

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Graphical verification by plotting {{{y=1/(x^2-x)}}} and {{{y=1/x}}}
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{{{drawing(300,300,-2,4,-2,4,grid(1),circle(2,1/2,0.12),graph(300,300,-2,4,-2,4,1/(x^2-x),1/x))}}}