Question Number 20116 by Tinkutara last updated on 22/Aug/17 $$\mathrm{If}\:{a}\:\mathrm{and}\:{b}\:\left(\neq\:\mathrm{0}\right)\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{equation}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:=\:\mathrm{0},\:\mathrm{then}\:\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{least}\:\mathrm{value}\:\mathrm{of}\:{x}^{\mathrm{2}} \:+\:{ax}\:+\:{b}\:\left({x}\:\in\:{R}\right). \\ $$ Answered by ajfour last updated on 22/Aug/17…
Question Number 20115 by Tinkutara last updated on 22/Aug/17 $$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{for}\:\mathrm{which}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\left(\mathrm{1}\:−\:{a}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:+\:\mathrm{2}{ax}\:−\:\mathrm{1}\:=\:\mathrm{0}\:\mathrm{has}\:\mathrm{roots} \\ $$$$\mathrm{belonging}\:\mathrm{to}\:\left(\mathrm{0},\:\mathrm{1}\right)\:\mathrm{is} \\ $$ Answered by ajfour last updated on 22/Aug/17…
Question Number 85648 by sakeefhasan05@gmail.com last updated on 23/Mar/20 $$\int\frac{\left(\mathrm{x}^{\mathrm{3}} −\mathrm{4}\right)}{\left(\mathrm{x}+\mathrm{1}\right)}\mathrm{dx} \\ $$ Answered by petrochengula last updated on 23/Mar/20 $$=\int\frac{{x}^{\mathrm{3}} +\mathrm{1}−\mathrm{5}}{\left({x}+\mathrm{1}\right)}{dx} \\ $$$$=\int\frac{{x}^{\mathrm{3}} +\mathrm{1}}{{x}+\mathrm{1}}{dx}−\mathrm{5}\int\frac{\mathrm{1}}{{x}+\mathrm{1}}{dx}…
Question Number 85649 by Hassen_Timol last updated on 23/Mar/20 $$\mathrm{Proove}\:\mathrm{that}\:: \\ $$$$ \\ $$$$\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}\:=\:\frac{\sqrt{\mathrm{6}}−\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$ Commented by mr W last updated on 23/Mar/20 $$\frac{\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}}=\frac{\sqrt{\mathrm{8}−\mathrm{4}\sqrt{\mathrm{3}}}}{\mathrm{4}}=\frac{\sqrt{\left(\sqrt{\mathrm{6}}\right)^{\mathrm{2}}…
Question Number 151181 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}\:\:\mid\boldsymbol{\mathrm{x}}\mid<\mathrm{1} \\ $$$$\mathrm{find}\:\:\mathrm{x}−\mathrm{4x}^{\mathrm{2}} +\mathrm{9x}^{\mathrm{3}} −\mathrm{16x}^{\mathrm{4}} +… \\ $$ Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\mathrm{S}\left({x}\right)\:=\:−\underset{{n}=\mathrm{1}}…
Question Number 85646 by M±th+et£s last updated on 23/Mar/20 $${show}\:{that} \\ $$$$\int\frac{\mathrm{1}}{\left[{x}\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)…\left({x}−{m}\right)\right]^{\mathrm{2}} }{dx}= \\ $$$$=\frac{\mathrm{1}}{\left({m}!\right)^{\mathrm{2}} }\underset{{n}=\mathrm{0}} {\overset{{m}} {\sum}}\frac{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} }{{n}−{x}}+\frac{\mathrm{2}}{\left({m}!\right)^{\mathrm{2}} }{ln}\mid\underset{{n}=\mathrm{0}} {\overset{{m}} {\prod}}\left({x}−{n}\right)^{\begin{pmatrix}{{m}}\\{{n}}\end{pmatrix}^{\mathrm{2}} \left({H}_{{m}−{n}} −{H}_{{n}} \right)}…
Question Number 20110 by mondodotto@gmail.com last updated on 22/Aug/17 Commented by ajfour last updated on 22/Aug/17 $$\mathrm{74658} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 151182 by mnjuly1970 last updated on 18/Aug/21 $$ \\ $$$$\int_{\mathrm{0}} ^{\:\infty} \frac{{sin}\left(\mathrm{2}{x}\right){ln}\left({x}\right)}{{x}}\:{dx}=\:{m}.\int_{\mathrm{0}} ^{\:\infty} \frac{\:{ln}\left(\mathrm{1}+\mathrm{2}{x}+{x}^{\mathrm{2}} \right)}{{x}\left({ln}^{\mathrm{2}} \left({x}\right)+\:\pi^{\:\mathrm{2}} \right)}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:{m}=?…. \\ $$ Terms of…
Question Number 151179 by mathdanisur last updated on 18/Aug/21 $$\mathrm{if}\:\:\mid\boldsymbol{\mathrm{x}}\mid<\mathrm{1} \\ $$$$\mathrm{find}\:\:\mathrm{x}+\mathrm{2x}^{\mathrm{2}} +\mathrm{3x}^{\mathrm{3}} +… \\ $$ Answered by Olaf_Thorendsen last updated on 18/Aug/21 $$\mathrm{S}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty}…
Question Number 85641 by abdomathmax last updated on 23/Mar/20 $${calculate}\:{A}_{\lambda} =\int_{\mathrm{3}} ^{\infty} \:\:\frac{{dx}}{\left({x}+\lambda\right)^{\mathrm{3}} \left({x}−\mathrm{2}\right)^{\mathrm{4}} }\:\:\:\left(\lambda>\mathrm{0}\right) \\ $$ Commented by mathmax by abdo last updated on…