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Question-38477




Question Number 38477 by Sr@2004 last updated on 26/Jun/18
Answered by $@ty@m last updated on 26/Jun/18
((a−b)/c)+((b−c)/a)+((c+a)/b)=1  ⇒((a−b)/c)+((c+a)/b)=1−((b−c)/a)  ⇒((b(a−b)+c(c+a))/(bc))=((a−b+c)/a)  ⇒(((a−b+c)(b+c))/(bc))=((a−b+c)/a)  ⇒((b+c)/(bc))=(1/a)  ⇒(1/b)+(1/c)=(1/a)
$$\frac{{a}−{b}}{{c}}+\frac{{b}−{c}}{{a}}+\frac{{c}+{a}}{{b}}=\mathrm{1} \\ $$$$\Rightarrow\frac{{a}−{b}}{{c}}+\frac{{c}+{a}}{{b}}=\mathrm{1}−\frac{{b}−{c}}{{a}} \\ $$$$\Rightarrow\frac{{b}\left({a}−{b}\right)+{c}\left({c}+{a}\right)}{{bc}}=\frac{{a}−{b}+{c}}{{a}} \\ $$$$\Rightarrow\frac{\left({a}−{b}+{c}\right)\left({b}+{c}\right)}{{bc}}=\frac{{a}−{b}+{c}}{{a}} \\ $$$$\Rightarrow\frac{{b}+{c}}{{bc}}=\frac{\mathrm{1}}{{a}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}=\frac{\mathrm{1}}{{a}} \\ $$

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