4/1 - x + 2/(1 - x)^2

I have the answer

6-4x/(x-1)(x-1)

but i dont understand how to get there

anyhelp is greatly appreciated (Happy)

thank you

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- Apr 7th 2009, 05:26 PMmadmattHelp? i dont get it
4/1 - x + 2/(1 - x)^2

I have the answer

6-4x/(x-1)(x-1)

but i dont understand how to get there

anyhelp is greatly appreciated (Happy)

thank you

- Apr 7th 2009, 05:31 PMProve It
$\displaystyle \frac{4}{1 - x} + \frac{2}{(1 - x)^2}$.

You need these terms to have the same denominator in order to add the numerators. To do this, multiply the first term by a cleverly disguised 1, in this case, $\displaystyle \frac{1 - x}{1 - x}$.

So $\displaystyle \frac{4}{1 - x} + \frac{2}{(1 - x)^2} = \frac{4}{1 - x}\times \frac{1 - x}{1 - x} + \frac{2}{(1 - x)^2}$

$\displaystyle = \frac{4(1 - x)}{(1 - x)^2} + \frac{2}{(1 - x)^2}$

$\displaystyle = \frac{4(1 - x) + 2}{(1 - x)^2}$

$\displaystyle = \frac{4 - 4x + 2}{(1 - x)^2}$

$\displaystyle = \frac{6 - 4x}{(1 - x)^2}$.

Also note that $\displaystyle (1 - x)^2 = (1 - x)(1 - x) = -(x - 1)(1 - x) = -[-(x - 1)(x - 1)] = (x - 1)(x - 1)$. - Apr 7th 2009, 05:35 PMmadmatt
(Bow)That helps a lot,

thank you very much