Question Number 77754 by abdomathmax last updated on 09/Jan/20 $${U}_{{n}} {isa}\:{sequence}\:{woch}\:{verify} \\ $$$${U}_{{n}} +{U}_{{n}+\mathrm{1}} ={n}^{\mathrm{2}} \left(−\mathrm{1}\right)^{{n}} \:\:\forall\:{n}\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{detdrmine}\:{U}_{{n}} \:{interm}\:{of}\:{n} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{nsture}\:{of}\:{the}\:{serie}\:\Sigma\:\frac{{U}_{{n}} }{{n}^{\mathrm{4}} } \\…
Question Number 77755 by abdomathmax last updated on 09/Jan/20 $$\left.\mathrm{1}\right){calculste}\:\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −{x}+{a}}}\:\:{with}\:\:{a}\:>\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left({a}\right)\:{at}\:{form}\:{of}\:{integral}\:{then}\:\:{find} \\ $$$${its}\:{value}. \\ $$$$ \\ $$$$ \\ $$ Commented…
Question Number 12218 by tawa last updated on 16/Apr/17 $$\mathrm{5}!!\:=\:? \\ $$ Commented by tawa last updated on 16/Apr/17 $$\mathrm{How}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{15} \\ $$ Commented by prakash…
Question Number 77752 by abdomathmax last updated on 09/Jan/20 $${let}\:{f}\left(\lambda\right)\:=\int_{−\infty} ^{+\infty} \:\frac{{sin}\left(\:\lambda{e}^{{x}} \:+{e}^{−{x}} \right)}{{x}^{\mathrm{2}} \:+\lambda^{\mathrm{2}} }{dx}\:{with}\:\lambda\geqslant\mathrm{0} \\ $$$$\left.\mathrm{1}\right)\:{detdrmine}\:{a}\:{explicit}\:{form}\:{of}\:{f}\left(\lambda\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{f}^{'} \left(\lambda\right)\:{at}\:{form}\:{ofintergral}\:{and}\:{find} \\ $$$${its}\:{value}. \\ $$…
Question Number 143284 by mathdanisur last updated on 12/Jun/21 Answered by mitica last updated on 12/Jun/21 Commented by loveineq last updated on 12/Jun/21 $$\mathrm{Very}\:\mathrm{nice}\:\mathrm{solution}. \\…
Question Number 77751 by abdomathmax last updated on 09/Jan/20 $${calculate}\:\int_{−\infty} ^{+\infty} \:\frac{{cos}\left({e}^{{x}} +{e}^{−{x}} \right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$ Commented by msup trace by abdo last…
Question Number 143287 by enter last updated on 12/Jun/21 Answered by Huy last updated on 12/Jun/21 $$\mathrm{Dieu}\:\mathrm{kien}:\:\mathrm{x}\geqslant−\mathrm{1} \\ $$$$\mathrm{Phuong}\:\mathrm{trinh}\:\mathrm{tuong}\:\mathrm{duong}: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2}\right)=\mathrm{5}\sqrt{\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}\right)} \\ $$$$<=>\mathrm{2}\left(\mathrm{x}^{\mathrm{2}}…
Question Number 77746 by mr W last updated on 09/Jan/20 $${Master}\:{comes}\:{after}\:{Chess}\:{disappeared}. \\ $$$${Boyka}\:{comes}\:{after}\:{Master}\:{disappeared}. \\ $$$${BK}\:{comes}\:{after}\:{Boyka}\:{disappeared}. \\ $$$${What}\:{comes}\:{after}\:{BK}\:{disappeared}? \\ $$$$\mathrm{1}.\:\:{A}−{Team} \\ $$$$\mathrm{2}.\:\:{Girlka} \\ $$$$\mathrm{3}.\:\:{Bezirksschornsteinfegermeister} \\ $$$$\mathrm{4}.\:\:{none}\:{from}\:{above}…
Question Number 12209 by tawa last updated on 16/Apr/17 $$\mathrm{For}\:\mathrm{all}\:\mathrm{n}\:\geqslant\:\mathrm{1}\:,\:\:\mathrm{n}\:\in\:\mathrm{Z},\:\:\mathrm{prove}\:\mathrm{that},\: \\ $$$$\mathrm{p}\left(\mathrm{n}\right)\::\:\mathrm{4}\:+\:\mathrm{8}\:+\:…\:+\:\mathrm{4n}\:=\:\mathrm{2n}\left(\mathrm{n}\:+\:\mathrm{1}\right) \\ $$ Commented by tawa last updated on 16/Apr/17 $$\mathrm{please}\:\mathrm{show}\:\mathrm{me}\:\mathrm{full}\:\mathrm{workings}\:\mathrm{sirs}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{all}. \\ $$ Answered…
Question Number 77745 by mathocean1 last updated on 09/Jan/20 $$\mathrm{ABC}\:\mathrm{is}\:\mathrm{any}\:\mathrm{triangle}. \\ $$$$\mathrm{C}'\:.\:\mathrm{B}'\:\:.\mathrm{A}'\:\:\mathrm{are}\:\mathrm{respectively}\:\mathrm{middles} \\ $$$$\mathrm{of}\:\left[\mathrm{AB}\right]\:.\:\left[\mathrm{AC}\right]\:\:\mathrm{and}\:\:\left[\mathrm{BC}\right]. \\ $$$$\mathrm{we}\:\mathrm{suppose}\:\mathrm{that}\: \\ $$$$\mathrm{AB}=\mathrm{c}\:\:\:\mathrm{AC}=\mathrm{b}\:\:\:\:\mathrm{BC}=\mathrm{a}. \\ $$$$\left.\mathrm{1}\right)\:\overset{\rightarrow\:} {\mathrm{u}}=\mathrm{a}^{\mathrm{2}} \mathrm{B}\overset{\rightarrow} {\mathrm{C}}+\mathrm{b}^{\mathrm{2}\:} \overset{\rightarrow} {\mathrm{C}A}+\mathrm{c}^{\mathrm{2}}…