Question Number 36843 by Penguin last updated on 06/Jun/18 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left({x}^{{x}^{{x}^{…..} } } \right) \\ $$ Commented by maxmathsup by imad last updated on 06/Jun/18…
Question Number 36828 by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18 Commented by tanmay.chaudhury50@gmail.com last updated on 06/Jun/18 $${these}\:{are}\:{useful}\:{formula}\:{hence}\:{posted}…{not}\: \\ $$$${question} \\ $$$$ \\ $$ Commented by…
Question Number 167884 by mathlove last updated on 28/Mar/22 $$\int{cosx}\:{dx}−\int{cosx}\:{dx}=? \\ $$ Answered by ArielVyny last updated on 28/Mar/22 $${seriously}?? \\ $$ Answered by Mathspace…
Question Number 167877 by infinityaction last updated on 28/Mar/22 Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 167879 by mnjuly1970 last updated on 28/Mar/22 $$ \\ $$$$\:\:\:\:{solve}\:{in}\:\:\mathbb{R} \\ $$$$ \\ $$$$\:\:\sqrt{{x}^{\:\mathrm{2}} −\mathrm{2}{x}\:+\mathrm{5}}\:+\:\sqrt{\mathrm{2}{x}^{\:\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{8}}\:=\mathrm{7} \\ $$$$ \\ $$$$\:\:\:\:\:\:−−−−− \\ $$ Commented…
Question Number 167860 by infinityaction last updated on 27/Mar/22 Commented by infinityaction last updated on 27/Mar/22 $${find}\:{the}\:{real}\:{roots}\: \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 167840 by mathlove last updated on 27/Mar/22 Commented by cortano1 last updated on 27/Mar/22 $$\:\:\begin{cases}{{f}\left({x}\right)=\mathrm{2}+\left({a}−\mathrm{1}\right){x}}\\{{g}\left({x}\right)=\mathrm{2}−\left({a}+\mathrm{1}\right){x}}\end{cases} \\ $$$$\:\:\:{f}\left({g}\left({x}\right)\right)−{g}\left({f}\left({x}\right)\right)={f}\left({a}−\mathrm{1}\right)+{g}\left({a}+\mathrm{1}\right) \\ $$$$\:\:\begin{cases}{{f}\left({g}\left({x}\right)\right)=\mathrm{2}+\left({a}−\mathrm{1}\right)\left\{\mathrm{2}−\left({a}+\mathrm{1}\right){x}\right\}}\\{{g}\left({f}\left({x}\right)\right)=\mathrm{2}−\left({a}+\mathrm{1}\right)\left\{\mathrm{2}+\left({a}−\mathrm{1}\right){x}\right\}}\end{cases} \\ $$$$\:\begin{cases}{{f}\left({g}\left({x}\right)\right)=\mathrm{2}{a}−\left({a}^{\mathrm{2}} −\mathrm{1}\right){x}}\\{{g}\left({f}\left({x}\right)\right)=−\mathrm{2}{a}−\left({a}^{\mathrm{2}} −\mathrm{1}\right){x}}\end{cases}…
Question Number 167839 by mathlove last updated on 27/Mar/22 Answered by mr W last updated on 27/Mar/22 $${say}\:{x}+{y}+{z}−\mathrm{3}={s}\: \\ $$$$\Rightarrow{x}+{y}+{z}−{s}=\mathrm{3} \\ $$$$\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{668}}}&{\frac{\mathrm{1}}{\mathrm{669}}}&{\frac{\mathrm{1}}{\mathrm{670}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{670}}}&{\frac{\mathrm{1}}{\mathrm{671}}}&{\frac{\mathrm{1}}{\mathrm{672}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{674}}}&{\frac{\mathrm{1}}{\mathrm{675}}}&{\frac{\mathrm{1}}{\mathrm{676}}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\\{{s}}\end{pmatrix}=\begin{pmatrix}{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{1}}\\{\mathrm{3}}\end{pmatrix} \\ $$$${s}=\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{668}}}&{\frac{\mathrm{1}}{\mathrm{669}}}&{\frac{\mathrm{1}}{\mathrm{670}}}&{\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{670}}}&{\frac{\mathrm{1}}{\mathrm{671}}}&{\frac{\mathrm{1}}{\mathrm{672}}}&{\mathrm{1}}\\{\frac{\mathrm{1}}{\mathrm{674}}}&{\frac{\mathrm{1}}{\mathrm{675}}}&{\frac{\mathrm{1}}{\mathrm{676}}}&{\mathrm{1}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{\mathrm{3}}\end{bmatrix}/\begin{bmatrix}{\frac{\mathrm{1}}{\mathrm{668}}}&{\frac{\mathrm{1}}{\mathrm{669}}}&{\frac{\mathrm{1}}{\mathrm{670}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{670}}}&{\frac{\mathrm{1}}{\mathrm{671}}}&{\frac{\mathrm{1}}{\mathrm{672}}}&{\mathrm{0}}\\{\frac{\mathrm{1}}{\mathrm{674}}}&{\frac{\mathrm{1}}{\mathrm{675}}}&{\frac{\mathrm{1}}{\mathrm{676}}}&{\mathrm{0}}\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}&{−\mathrm{1}}\end{bmatrix}=\mathrm{2012} \\…
Question Number 167821 by mathlove last updated on 26/Mar/22 Commented by dangduomg last updated on 26/Mar/22 $$\Rightarrow\:{E}\:=\:\left(\mathrm{4}^{\mathrm{1}/\mathrm{5}} \right)^{\left(\mathrm{5}^{\mathrm{1}/\mathrm{4}} \right)^{{E}} } \\ $$$$=\:\mathrm{4}^{\mathrm{1}/\mathrm{5}×\left(\mathrm{5}^{\mathrm{1}/\mathrm{4}} \right)^{{E}} } \\…
Question Number 102278 by I want to learn more last updated on 08/Jul/20 $$\mathrm{Find}\:\mathrm{the}\:\mathrm{greatest}\:\mathrm{coefficient}\:\mathrm{in}\:\mathrm{the}\:\mathrm{following}\:\mathrm{without} \\ $$$$\mathrm{actually}\:\mathrm{expand}. \\ $$$$\left(\mathrm{i}\right)\:\:\:\:\:\:\:\:\left(\mathrm{5}\:\:−\:\:\mathrm{3x}\right)^{\mathrm{10}} \\ $$$$\left(\mathrm{ii}\right)\:\:\:\:\:\:\:\:\left(\mathrm{5}\:\:+\:\:\mathrm{3x}\right)^{−\:\mathrm{10}} \\ $$ Commented by mr…