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 February 18th, 2014, 05:41 AM #1 Newbie   Joined: Feb 2014 Posts: 6 Thanks: 0 What is the simplest form of this What is the simplest form of this? $\frac{m^2}{(m^3-m)}+\frac{1}{(2-2m)}=$
 February 18th, 2014, 07:43 AM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,788 Thanks: 970 Re: What is the simplest form of this 1 / [2(m+1)] 1st step: 1 / (2 - 2m) = 1 / [2(1 - m)] = - 1 / [2(m - 1)] ; make sure you understand that...
March 9th, 2014, 01:20 AM   #3
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Re: What is the simplest form of this

Quote:
 Originally Posted by naufalzhafran What is the simplest form of this? $\frac{m^2}{(m^3-m)}+\frac{1}{(2-2m)}=$
$\frac{m^{2}}{m^{3}-m}+\frac{1}{2-2m}=\frac{m^{2}}{m(m^{2}-1)}-\frac{1}{2}\left (\frac{1}{m-1} \right )$

$=\frac{m^{2}}{m(m-1)(m+1)}-\frac{1}{2}\left (\frac{1}{m-1} \right )$

$=\frac{1}{m-1}\left ( \frac{m^{2}}{m(m+1)}-\frac{1}{2} \right )=\frac{1}{m-1}\left ( \frac{2m^{2}-(m^2+m)}{2(m^2+m)} \right )$

$=\frac{1}{m-1}\left ( \frac{m^2-m}{2m(m+1)} \right )=\frac{1}{m-1}\left (\frac{m(m-1)}{2m(m+1)} \right )=\frac{1}{2(m+1)}$

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