Question Number 70768 by abdusalamyussif@gmail.com last updated on 07/Oct/19
Commented by abdusalamyussif@gmail.com last updated on 07/Oct/19
$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much}\:\mathrm{sir} \\ $$
Answered by MJS last updated on 07/Oct/19
$$\frac{{m}\left({m}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} −\mathrm{2}{m}^{\mathrm{2}} \left({m}^{\mathrm{2}} +\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\left({m}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{1}} }= \\ $$$$={m}\left({m}^{\mathrm{2}} +\mathrm{1}\right)\left(\left({m}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} −\frac{\mathrm{2}{m}}{\left({m}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} }\right)= \\ $$$$={m}\left({m}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left({m}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{m}\right)= \\ $$$$={m}\left({m}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left({m}−\mathrm{1}\right)^{\mathrm{2}} = \\ $$$$={m}\left({m}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{{m}^{\mathrm{2}} +\mathrm{1}} \\ $$
Commented by abdusalamyussif@gmail.com last updated on 07/Oct/19
$$\mathrm{thanks}\:\mathrm{so}\:\mathrm{much} \\ $$