Answer :
I have not done this in advance. I'm just going to write it down
and see what I can do with it:
[ 1/(x+3)² - 1/x² ] / 3
Multiply the top and bottom by (x+3)² :
[ 1 - (x+3)²/x² ] / 3 (x+3)²
Multiply the top and bottom by x² :
[ x² - (x+3)² ] / 3 x² (x+3)²
Now it's just a matter of expanding and cleaning things up,
and hope and pray that a lot of things cancel.
Eliminate the parentheses on top and bottom:
[ x² - x² - 6x -9 ] / 3 x² (x² + 6x + 9)
Combine the x² terms on top, and divide top and bottom by 3 :
[ - 2x - 3 ] / x² (x² + 6x + 9)
Finally, all I can make of this is:
- (2x + 3) / [ x(x+3) ]² .
That's not a whole lot prettier than the original form, but at least
we got rid of those fractions in the numerator of a fraction.
I hope this is some help to you.
and see what I can do with it:
[ 1/(x+3)² - 1/x² ] / 3
Multiply the top and bottom by (x+3)² :
[ 1 - (x+3)²/x² ] / 3 (x+3)²
Multiply the top and bottom by x² :
[ x² - (x+3)² ] / 3 x² (x+3)²
Now it's just a matter of expanding and cleaning things up,
and hope and pray that a lot of things cancel.
Eliminate the parentheses on top and bottom:
[ x² - x² - 6x -9 ] / 3 x² (x² + 6x + 9)
Combine the x² terms on top, and divide top and bottom by 3 :
[ - 2x - 3 ] / x² (x² + 6x + 9)
Finally, all I can make of this is:
- (2x + 3) / [ x(x+3) ]² .
That's not a whole lot prettier than the original form, but at least
we got rid of those fractions in the numerator of a fraction.
I hope this is some help to you.
[tex]\frac{\frac{1}{(x+3)^2}-\frac{1}{x^2}}{3}=\frac{1}{3(x+3)^2}-\frac{1}{3x^2}=\frac{x^2}{3x^2(x+3)^2}-\frac{(x+3)^2}{3x^2(x+3)^2}=\frac{x^2-x^2-6x-9}{3x^3(x+3)^2}\\\\=\frac{-6x-9}{3x^2(x+3)^2}=\frac{3(-2x-3)}{3x^2(x+3)^2}=-\frac{2x+3}{x^2(x+3)^2}[/tex]
