Question Number 46681 by Aditya789 last updated on 30/Oct/18
Commented by math1967 last updated on 30/Oct/18
$${is}\:{it}\:{solve}? \\ $$
Commented by Aditya789 last updated on 30/Oct/18
$$\mathrm{yes} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Oct/18
$$\frac{{x}−{b}−{c}}{{a}}−\mathrm{1}+\frac{{x}−{c}−{a}}{{b}}−\mathrm{1}+\frac{{x}−{a}−{b}\left(\leftarrow{should}\:{be}\right)}{{c}}−\mathrm{1}=\mathrm{0} \\ $$$$\frac{{x}−{a}−{b}−{c}}{{a}}+\frac{{x}−{a}−{b}−{c}}{{b}}+\frac{{x}−{a}−{b}−{c}}{{c}}=\mathrm{0} \\ $$$$\left({x}−{a}−{b}−{c}\right)\left(\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}\right)=\mathrm{0} \\ $$$${so}\:{x}−{a}−{b}−{c}=\mathrm{0} \\ $$$${x}={a}+{b}+{c} \\ $$$$\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}{can}\:{not}\:{be}\:{equals}\:{to}\:{zero} \\ $$$$ \\ $$