Question 134730
I assume this is the problem:
{{{1/3-5/(3x)+2/x^2}}}
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{{{1/3 - 3/x^2}}}
We want to get fractions over a common denominator; it will be 3x^2 on both:
{{{((x^2 + 5x + 3(2)))/(3x^2)}}}
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{{{((x^2 - 3(3)))/(3x^2)}}}
which is
{{{((x^2 + 5x + 6))/(3x^2)}}}
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{{{((x^2 - 9))/(3x^2)}}}
:
Invert the dividing fraction and multiply:
{{{((x^2 + 5x + 6))/(3x^2)}}} * {{{(3x^2)/((x^2 - 9))}}}
:
You can see the 3x^2's will cancel leaving us with:
{{{((x^2 + 5x + 6))/((x^2 - 9))}}}
:
Both denominator and numerator will factor:
{{{((x+2)(x+3))/((x-3)(x+3)))}}}
:
Cancel out (x+3) and you have:
{{{((x+2))/((x-3))}}}
:
How about this? Did it make sense to you?