quick show question (1 Viewer)

Trebla · Apr 5, 2010

It doesn't matter. There is no difference to proving

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

compared to

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

If you're still worried about it, then simply do the steps you did from RHS to LHS backwards to go from LHS to RHS.

$\dfrac{1}{x^2-9} = \dfrac{1}{6}\:\dfrac{6}{x^2-9}\\\\= \dfrac{1}{6}\:\dfrac{(x+3)-(x-3)}{(x-3)(x+3)}\\\\=\dfrac{1}{6}\left(\dfrac{1}{x-3}-\dfrac{1}{x+3}\right)$

cutemouse · Apr 6, 2010

thongetsu said:
mmmm showing/proving is one of the easiest maths questions considering you are given the answer

Not in Ext 2 lol...

And yes LHS=RHS is the same as RHS=LHS.

thongetsu · Apr 6, 2010

fortunately i dont do ext.2

quick show question (1 Viewer)

Trebla

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cutemouse

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thongetsu

Where aren't I?

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