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Question Number 19268 by khamizan833@yahoo.com last updated on 08/Aug/17

Answered by ajfour last updated on 08/Aug/17

$$\mathrm{f}\left(\mathrm{x}\right)=\mathrm{x}(\frac{1}{1-2^{\mathrm{x}}}-\frac{1}{2})$$$$\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}(-\mathrm{x})=\mathrm{x}[(\frac{1}{1-2^{\mathrm{x}}}-\frac{1}{2})-(\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}-1}-\frac{1}{2})]$$$$=-\frac{(2^{\mathrm{x}}+1)\mathrm{x}}{(2^{\mathrm{x}}-1)}$$$$\Rightarrow \mathrm{f}\left(5\right)+\mathrm{f}(-5)=-\frac{33\times 5}{31}=-\frac{165}{31}$$$$\mathrm{f}\left(\mathrm{x}\right)-\mathrm{f}(-\mathrm{x})=\mathrm{x}[(\frac{1}{1-2^{\mathrm{x}}}-\frac{1}{2})+(\frac{2^{\mathrm{x}}}{2^{\mathrm{x}}-1}-\frac{1}{2})]$$$$=0$$$$\Rightarrow \mathrm{f}\left(3\right)-\mathrm{f}(-3)=0$$$$\mathrm{so},\frac{\mathrm{f}\left(3\right)-\mathrm{f}(-3)}{\mathrm{f}\left(5\right)+\mathrm{f}(-5)}=0.$$

Commented by khamizan833@yahoo.com last updated on 08/Aug/17

$$\mathrm{thank}\mathrm{you}\mathrm{very}\mathrm{much},\mathrm{sir}.$$

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