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Question Number 116516 by sandy_delta last updated on 04/Oct/20

Answered by som(math1967) last updated on 04/Oct/20

$$(\frac{1}{\mathrm{a}+\mathrm{b}}+\frac{1}{\mathrm{b}+\mathrm{c}}+\frac{1}{\mathrm{c}+\mathrm{a}})(\mathrm{a}+\mathrm{b}+\mathrm{c})=(\mathrm{a}+\mathrm{b}+\mathrm{c})\times \frac{7}{10}$$$$\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}}+\frac{\mathrm{a}+\mathrm{b}+c}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{c}+\mathrm{a}}=\frac{49}{10}$$$$[\because \mathrm{a}+\mathrm{b}+\mathrm{c}=7]$$$$1+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}+1+\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+1+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}=\frac{49}{10}$$$$\therefore \frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\frac{49}{10}-3=\frac{19}{10}\mathrm{ans}$$

Commented by sandy_delta last updated on 04/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{Sir}$$

Answered by bobhans last updated on 05/Oct/20

$$\frac{\mathrm{a}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{b}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\mathrm{s}$$$$\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{b}+\mathrm{c}}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{c}}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{\mathrm{a}+\mathrm{b}}=\mathrm{s}+3$$$$\frac{7}{\mathrm{b}+\mathrm{c}}+\frac{7}{\mathrm{a}+\mathrm{c}}+\frac{7}{\mathrm{a}+\mathrm{b}}=\mathrm{s}+3$$$$7(\frac{1}{\mathrm{b}+\mathrm{c}}+\frac{1}{\mathrm{a}+\mathrm{c}}+\frac{1}{\mathrm{a}+\mathrm{b}})=\mathrm{s}+3$$$$\Rightarrow \mathrm{s}=7\left(\frac{7}{10}\right)-3=\frac{49}{10}-\frac{30}{10}=\frac{19}{10}$$

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