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Question Number 60370 by ANTARES VY last updated on 20/May/19

Answered by math1967 last updated on 20/May/19

0

$$\mathrm{0} \\ $$

Answered by tanmay last updated on 20/May/19

N_r =(a^2 /(b+c))+a+(b^2 /(a+c))+b+(c^2 /(a+b))+c−(a+b+c)  ((a^2 +ab+ac)/(b+c))+((b^2 +ab+bc)/(a+c))+((c^2 +ca+cb)/(a+b))−(a+b+c)  =(a/(b+c))×(a+b+c)+(b/(a+c))×(a+b+c)+(c/(a+b))×(a+b+c)−(a+b+c)  =(a+b+c)((a/(b+c))+(b/(a+c))+(c/(a+b))−1)  =0  so given problem  =(N_r /D_r )=(0/(a+b+c))=0

$${N}_{{r}} =\frac{{a}^{\mathrm{2}} }{{b}+{c}}+{a}+\frac{{b}^{\mathrm{2}} }{{a}+{c}}+{b}+\frac{{c}^{\mathrm{2}} }{{a}+{b}}+{c}−\left({a}+{b}+{c}\right) \\ $$$$\frac{{a}^{\mathrm{2}} +{ab}+{ac}}{{b}+{c}}+\frac{{b}^{\mathrm{2}} +{ab}+{bc}}{{a}+{c}}+\frac{{c}^{\mathrm{2}} +{ca}+{cb}}{{a}+{b}}−\left({a}+{b}+{c}\right) \\ $$$$=\frac{{a}}{{b}+{c}}×\left({a}+{b}+{c}\right)+\frac{{b}}{{a}+{c}}×\left({a}+{b}+{c}\right)+\frac{{c}}{{a}+{b}}×\left({a}+{b}+{c}\right)−\left({a}+{b}+{c}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left(\frac{{a}}{{b}+{c}}+\frac{{b}}{{a}+{c}}+\frac{{c}}{{a}+{b}}−\mathrm{1}\right) \\ $$$$=\mathrm{0} \\ $$$${so}\:{given}\:{problem}\:\:=\frac{{N}_{{r}} }{{D}_{{r}} }=\frac{\mathrm{0}}{{a}+{b}+{c}}=\mathrm{0} \\ $$$$ \\ $$

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