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Question-70768




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

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