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Question-38696




Question Number 38696 by Sr@2004 last updated on 28/Jun/18
Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18
(1/(b+c))+(1/(c+a))=(2/(a+b))  (1/(b+c))−(1/(a+b))=(1/(a+b))−(1/(c+a))  ((a+b−b−c)/((a+b)(b+c)))=((c+a−a−b)/((a+b)(c+a)))  ((a−c)/(b+c))=((c−b)/(c+a))  a^2 −c^2 =c^2 −b^2   a^2 +b^2 =2c^2 proved  pls correct the question
$$\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}=\frac{\mathrm{2}}{{a}+{b}} \\ $$$$\frac{\mathrm{1}}{{b}+{c}}−\frac{\mathrm{1}}{{a}+{b}}=\frac{\mathrm{1}}{{a}+{b}}−\frac{\mathrm{1}}{{c}+{a}} \\ $$$$\frac{{a}+{b}−{b}−{c}}{\left({a}+{b}\right)\left({b}+{c}\right)}=\frac{{c}+{a}−{a}−{b}}{\left({a}+{b}\right)\left({c}+{a}\right)} \\ $$$$\frac{{a}−{c}}{{b}+{c}}=\frac{{c}−{b}}{{c}+{a}} \\ $$$${a}^{\mathrm{2}} −{c}^{\mathrm{2}} ={c}^{\mathrm{2}} −{b}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} =\mathrm{2}{c}^{\mathrm{2}} {proved} \\ $$$${pls}\:{correct}\:{the}\:{question} \\ $$$$ \\ $$

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