CreolaTeti359
74

# Need a step by step on how to solve this please

(2) Answers
bolt

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.

Nunia1567

$\frac{\frac{1}{(x+3)^2}-\frac{1}{x^2}}{3}=\frac{1}{3(x+3)^2}-\frac{1}{3x^2}=\frac{x^2}{3x^2(x+3)^2}-\frac{(x+3)^2}{3x^2(x+3)^2}=\frac{x^2-x^2-6x-9}{3x^3(x+3)^2}\\\\=\frac{-6x-9}{3x^2(x+3)^2}=\frac{3(-2x-3)}{3x^2(x+3)^2}=-\frac{2x+3}{x^2(x+3)^2}$

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