if x = (2ab+2bc+2ca)/(a+b+c), then prove that, [(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3 please help!!!
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Originally Posted by earthboy if x = (2ab+2bc+2ca)/(a+b+c), then prove that, [(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3 please help!!! is that menth to be $\displaystyle \frac{6abc}{(x-a)(x-b)(x-c)}$ or as it is written $\displaystyle \frac{6abc}{x-a}\times(x-b)(x-c)$
Quote: Originally Posted by earthboy if x = (2ab+2bc+2ca)/(a+b+c), then prove that, [(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3 please help!!! is that menth to be or as it is written i meant the first exp. not the second one
Originally Posted by earthboy if x = (2ab+2bc+2ca)/(a+b+c), then prove that, [(x+a)/(x-a)] + [(x+b)/(x-b)] +[(x+c)/(x-c)] - [6abc/(x-a)(x-b)(x-c)] = 3 start by combining the first three terms on the left with the last term using the common denominator $\displaystyle (x-a)(x-b)(x-c)$
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