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Question-116611




Question Number 116611 by gaminghanzo12 last updated on 05/Oct/20
Commented by Dwaipayan Shikari last updated on 05/Oct/20
Q 116516
$$\mathrm{Q}\:\mathrm{116516} \\ $$
Answered by TANMAY PANACEA last updated on 05/Oct/20
(a/(b+c))+1+(b/(a+c))+1+(c/(a+b))+1−3  (a+b+c)((1/(b+c))+(1/(a+c))+(1/(a+b)))−3  7×(7/(10))−3=((49−30)/(10))=((19)/(10))
$$\frac{{a}}{{b}+{c}}+\mathrm{1}+\frac{{b}}{{a}+{c}}+\mathrm{1}+\frac{{c}}{{a}+{b}}+\mathrm{1}−\mathrm{3} \\ $$$$\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{a}+{c}}+\frac{\mathrm{1}}{{a}+{b}}\right)−\mathrm{3} \\ $$$$\mathrm{7}×\frac{\mathrm{7}}{\mathrm{10}}−\mathrm{3}=\frac{\mathrm{49}−\mathrm{30}}{\mathrm{10}}=\frac{\mathrm{19}}{\mathrm{10}} \\ $$

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