Question 190524
Is the expression {{{(x+5/4)/(x^2+5x)}}}???



If so, then...



{{{(x+5/4)/(x^2+5x)}}} Start with the given expression



{{{(4x+cross(4)(5/cross(4)))/(4*x^2+4*5x)}}} Multiply EVERY term by the inner LCD 4 to clear out the inner fractions.



{{{(4x+5)/(4x^2+20x)}}} Multiply and simplify




So {{{(x+5/4)/(x^2+5x)}}} simplifies to {{{(4x+5)/(4x^2+20x)}}}



In short, {{{(x+5/4)/(x^2+5x)=(4x+5)/(4x^2+20x)}}} where {{{x<>-5}}} or {{{x<>0}}}



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OR...


Is the expression {{{((x+5)/4)/(x^2+5x)}}} ???



If so, then...





{{{((x+5)/4)/(x^2+5x)}}} Start with the given expression



{{{((x+5)/4)/((x^2+5x)/1)}}} Rewrite {{{x^2+5x}}} as {{{(x^2+5x)/1}}}



{{{((x+5)/4)(1/(x^2+5x))}}} Multiply the first fraction by the reciprocal of the second fraction.



{{{(x+5)/(4(x^2+5x))}}} Combine the fractions



{{{(x+5)/(4x(x+5))}}} Factor the denominator



{{{highlight((x+5))/(4x*highlight((x+5)))}}} Highlight the common terms.



{{{cross((x+5))/(4x*cross((x+5)))}}} Cancel out the common terms.



{{{1/(4x)}}} Simplify




So {{{((x+5)/4)/(x^2+5x)}}} simplifies to {{{1/(4x)}}}



In short, {{{((x+5)/4)/(x^2+5x)=1/(4x)}}} where {{{x<>-5}}} or {{{x<>0}}}