Question Number 184305 by HeferH last updated on 05/Jan/23 $${if}:\:{f}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:+\:\mathrm{2}{x} \\ $$$${find}\:{x}: \\ $$$$\:{f}\left({f}\left({f}\left({x}\:+\:\mathrm{2}\right)\right)\right)\:=\:\mathrm{99}\:\mathrm{999}\:\mathrm{999}\: \\ $$ Answered by Frix last updated on 05/Jan/23 $${x}={t}−\mathrm{3}…
Question Number 184304 by mathlove last updated on 05/Jan/23 Answered by Frix last updated on 05/Jan/23 $${a}=\frac{\mathrm{1}+\sqrt{\mathrm{4}{t}+\mathrm{221}}}{\mathrm{2}} \\ $$$${a}+\frac{\mathrm{2}}{{a}}=\mathrm{1}\:\Leftrightarrow\:{a}^{\mathrm{2}} −{a}+\mathrm{2}=\mathrm{0}\:\Leftrightarrow\:{t}=−\mathrm{57} \\ $$$${P}={t}^{\mathrm{2}} −\mathrm{1226}=\mathrm{2023} \\ $$…
Question Number 118770 by obaidullah last updated on 19/Oct/20 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =? \\ $$$${y}=\left(\frac{{n}!}{{n}^{{n}} }\right)^{\frac{\mathrm{1}}{{n}}} =\frac{\mathrm{1}}{{n}}{ln}\left(\frac{{n}!}{{n}^{{n}} }\right)=\frac{\mathrm{1}}{{n}}\left[{lnn}!−{nlnn}\right] \\ $$$$\Rightarrow\underset{{x}\rightarrow\infty} {\mathrm{lim}}{lny}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{{n}}\left[{nlnn}−{n}−{nlnn}\right] \\ $$$$\Rightarrow{ln}\underset{{x}\rightarrow\infty} {\mathrm{lim}}{y}=\underset{{x}\rightarrow\infty}…
Question Number 184302 by cortano1 last updated on 05/Jan/23 Answered by mr W last updated on 05/Jan/23 $${say}\:{f}\left({x}\right)={Ap}^{{x}} \\ $$$${Ap}^{{x}−\mathrm{1}} +{Ap}^{{x}+\mathrm{1}} =\sqrt{\mathrm{3}}{Ap}^{{x}} \\ $$$$\mathrm{1}+{p}^{\mathrm{2}} =\sqrt{\mathrm{3}}{p}…
Question Number 184280 by mr W last updated on 04/Jan/23 $$\:\:{Given}\:\begin{cases}{{a}_{\mathrm{0}} =\mathrm{1}}\\{{a}_{{n}+\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{3}{a}_{{n}} +\sqrt{\mathrm{5}{a}_{{n}} ^{\mathrm{2}} −\mathrm{4}}\:\right)}\end{cases} \\ $$$$\:\forall{n}\geqslant\mathrm{0}\:,\:{n}\in{I}\: \\ $$$$\:\:{find}\:{a}_{{n}} . \\ $$ Commented by…
Question Number 118740 by Jamshidbek2311 last updated on 19/Oct/20 $${f}\left({x}+\mathrm{2}\right)+{f}\left({x}−\mathrm{1}\right)=\mathrm{2}{x}^{\mathrm{2}} +\mathrm{14} \\ $$$${f}\left({x}\right)=? \\ $$ Commented by PRITHWISH SEN 2 last updated on 19/Oct/20 $$\mathrm{x}=\mathrm{x}+\mathrm{1}…
Question Number 184270 by mathlove last updated on 04/Jan/23 $$\sqrt[{\mathrm{6}}]{−\mathrm{64}}\centerdot\sqrt[{\mathrm{7}}]{−\mathrm{128}}=? \\ $$ Answered by HeferH last updated on 04/Jan/23 $$\sqrt[{\mathrm{6}}]{−\mathrm{1}\centerdot\mathrm{2}^{\mathrm{6}} }\:\centerdot\:\sqrt[{\mathrm{7}}]{−\mathrm{1}\centerdot\mathrm{2}^{\mathrm{7}} }\:=\:\mathrm{2}{i}\centerdot\mathrm{2}{i}\:=\:\mathrm{4}{i}^{\mathrm{2}} =−\mathrm{4} \\ $$$$\:{mistake}:\:\sqrt[{\mathrm{6}}]{−\mathrm{1}}\:,\:\sqrt[{\mathrm{7}}]{−\mathrm{1}}\:\neq\:{i}…
Question Number 184258 by universe last updated on 04/Jan/23 $$\:\:\:\mathrm{2yz}−\mathrm{4z}+\mathrm{2x}−\mathrm{2}=\mathrm{0} \\ $$$$\:\:\:\mathrm{2xz}−\mathrm{2z}+\mathrm{2y}−\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\mathrm{2xy}−\mathrm{4x}−\mathrm{2y}+\mathrm{2z}+\mathrm{4}=\mathrm{0} \\ $$$$\:\:\:\mathrm{how}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{all}\:\mathrm{values}\:\mathrm{of}\:\mathrm{the}\:\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)\:? \\ $$ Answered by SEKRET last updated on 04/Jan/23…
Question Number 118712 by 1549442205PVT last updated on 19/Oct/20 $$\mathrm{Prove}\:\mathrm{the}\:\mathrm{following}\:\mathrm{inequalities}: \\ $$$$\left.\mathrm{1}\right)\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{n}} >\mathrm{n}!\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} ,\mathrm{n}>\mathrm{1} \\ $$$$\left.\mathrm{2}\right)\mid\mathrm{sinnx}\mid\leqslant\mathrm{n}\mid\mathrm{sinx}\mid\:\mathrm{for}\:\forall\mathrm{n}\in\mathrm{N}^{\ast} \\ $$ Answered by Dwaipayan Shikari last updated on…
Question Number 184251 by Shrinava last updated on 04/Jan/23 $$\mathrm{If}\:\:\:\mathrm{x}\:\sqrt{\mathrm{y}}\:=\:\mathrm{2904} \\ $$$$\mathrm{Find}:\:\:\:\mathrm{y}=? \\ $$ Answered by SEKRET last updated on 04/Jan/23 $$\:\:\:\boldsymbol{\mathrm{x}}\centerdot\sqrt{\boldsymbol{\mathrm{y}}}\:=\:\mathrm{2}^{\mathrm{3}} \centerdot\mathrm{3}\centerdot\mathrm{11}^{\mathrm{2}} \\ $$$$\:\:\begin{cases}{\boldsymbol{\mathrm{x}}=\mathrm{2}}\\{\sqrt{\boldsymbol{\mathrm{y}}}\:=\mathrm{2}^{\mathrm{2}}…