Question 342768
No, not completely.
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The square root function needs a non-negative argument.
{{{((x-2)/(x-1))>=0}}}
Break up the number line into 3 regions.
Region 1:({{{-infinity}}},{{{1}}})
Region 2:({{{1}}},{{{2}}})
Region 3:({{{2}}},{{{infinity}}})
Choose a point in the region, not an endpoint.
Test the inequality.
If the inequality is satisfied, the region is part of the solution.
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Region 1: {{{x=0}}}
{{{(x-2)/(x-1)>=0}}}
{{{-2/-1>=0}}}
{{{2>=0}}}
True, this region is part of the solution. 
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Region 2: {{{x=3/2}}}
{{{(x-2)/(x-1)>=0}}}
{{{(3/2-2)/(3/2-1)>=0}}}
{{{-(1/2)/(1/2)>=0}}}
{{{-1>=0}}}
False, this region is not part of the solution.
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Region 3: {{{x=3}}}
{{{(x-2)/(x-1)>=0}}}
{{{(3-2)/(3-1)>=0}}}
{{{1/2>=0}}}
True, this region is part of the solution.
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Since {{{x=2}}} also solves the inequality, it is also included.
{{{x=1}}} is not included in any solution region because it leads to a division by zero.
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Domain:({{{-infinity}}},{{{1}}}) U [{{{2}}},{{{infinity}}})