Question 888046
You have a square root in a denominator, so multiplying the term with the square root in its denominator by
{{{(sqrt(x-5)/sqrt(x-5))}}} leads you to an equivalent expression without a square root in a denominator
As long as {{{x-5>=0}}}<--->{{{x>=5}}} the resulting expression is equivalent because
{{{(sqrt(x-5)/sqrt(x-5))=1}}} when {{{x-5>=0}}}<--->{{{x>=5}}} .
 
{{{3*sqrt(x-5)+4/sqrt(x-5)+1 = 3*sqrt(x-5)+(4/sqrt(x-5))*(sqrt(x-5)/sqrt(x-5))+1 = 3*sqrt(x-5)+4sqrt(x-5)/(sqrt(x-5))^2+1 = 3*sqrt(x-5)+4sqrt(x-5)/(x-5)+1 }}}
The last expression above can be transformed further, but it transforms into expressions that are not really simpler.
Some possibilities are:
{{{3*sqrt(x-5)+4*sqrt(x-5)/(x-5)+1}}}={{{(3+4/(x-5))sqrt(x-5)+1}}}={{{((3(x-5)+4)/(x-5))*sqrt(x-5)+1}}}={{{((3x-15+4)/(x-5))*sqrt(x-5)+1}}}={{{((3x-11)/(x-5))*sqrt(x-5)+1}}}={{{((3x-11)*sqrt(x-5)+x-5)/(x-5)}}}