Question 167398
1. simplify: 
:
{{{(4xy^3)/z^2}}}
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{{{((8x^2y)/z^3)^2}}}
:
Square the denominator fraction
{{{(4xy^3)/z^2}}}
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{{{(64x^4y^2)/z^6}}}
:
Invert the dividing fraction and multiply, cancel like terms
{{{(4xy^3)/z^2}}} * {{{z^6/(64x^4y^2)}}} = {{{(yz^4)/(16x^3)}}}
:

2. Simplify 
:
{{{d/((d^2-9))}}}+{{{5/((2d+6))}}}
:
Factor, then put over a common denominator:
{{{d/((d-3)(d+3))}}}+{{{5/(2(d+3))}}} = {{{(2d + (d-3)*5)/((2(d+3)(d-3)))}}} = {{{(2d + 5d - 15)/((2(d+3)(d-3)))}}} = {{{(7d - 15)/((2(d+3)(d-3)))}}}