Question 116948
1) 
{{{((x^2+7x+12))/((x+4))}}}*{{{1/((x+3))}}}
The first thing you want to do in these problems is see what can be factored
{{{((x+3)(x+4))/((x+4))}}}*{{{1/((x+3))}}} = 1
Note that you have (x+3) and (x+4) in the numerator and the denominator cancel these and that leaves 1
:
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2)
{{{25xy^2/(7z)}}}
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{{{5x^2y^2/(14z^2)}}}
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Remember, when you divide by fractions, you invert the dividing fraction & multiply
{{{25xy^2/(7z)}}} * {{{(14z^2)/(5x^2y^2)}}}
Cancel the following:
5 into 25
7 into 14
x in the numerator into x^2 of the denominator
y^2 into y^2
z in the denominator into z^2 in the numerator
Leaving us with:
5 * {{{(2z)/x}}} = {{{(10z)/x}}}
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How about this, is this the help you need? Any questions?