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Author: Tinku Tara

Question-197838

Question Number 197838 by hardmath last updated on 30/Sep/23 Answered by MM42 last updated on 30/Sep/23 $$\frac{\mathrm{2}^{\mathrm{49}} }{\mathrm{2}^{\mathrm{49}} +\mathrm{1}}+\frac{\mathrm{2}^{\mathrm{48}} }{\mathrm{2}^{\mathrm{48}} +\mathrm{1}}+…+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{48}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{49}} +\mathrm{1}} \\ $$$$=\mathrm{1}+\mathrm{1}+…+\mathrm{1}=\:\mathrm{49}\:\checkmark…

lim-x-2-cosx-1-cosx-

Question Number 197832 by mathlove last updated on 30/Sep/23 $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{2}+\sqrt{{cosx}}}{−\mathrm{1}+\sqrt{{cosx}}}=? \\ $$ Answered by MM42 last updated on 30/Sep/23 $${let}\:\:{f}\left({x}\right)=\frac{\mathrm{2}+\sqrt{{cosx}}}{−\mathrm{1}+\sqrt{{cosx}}} \\ $$$${for}\:\:{a}_{{n}} =\mathrm{2}{n}\pi\Rightarrow{lim}_{{n}\rightarrow\infty} \:{f}\left({a}_{{n}}…

I-2-6-x-1-x-1-dx-

Question Number 197802 by cortano12 last updated on 29/Sep/23 $$\:\:\:\mathrm{I}=\underset{−\mathrm{2}} {\overset{\mathrm{6}} {\int}}\:\frac{\mid\mathrm{x}−\mathrm{1}\mid}{\mathrm{x}−\mathrm{1}}\:\mathrm{dx}\:=? \\ $$ Answered by MM42 last updated on 29/Sep/23 $${I}=\int_{−\mathrm{2}} ^{\mathrm{1}} −{dx}+\int_{\mathrm{1}} ^{\mathrm{6}}…

Question-197795

Question Number 197795 by cortano12 last updated on 29/Sep/23 Answered by AST last updated on 29/Sep/23 $${Let}\:{y}={px};{x}\left({p}+\mathrm{1}\right)={x}^{\mathrm{2}} \left(\mathrm{1}+{p}^{\mathrm{2}} \right)\:;{x}\neq\mathrm{0}\Rightarrow{p}^{\mathrm{2}} ={p}\Rightarrow{p}=\mathrm{0}\:{or}\:\mathrm{1} \\ $$$${p}=\mathrm{0}\Rightarrow{y}=\mathrm{0}\Rightarrow{x}^{\mathrm{2}} ={x}\Rightarrow{x}=\mathrm{1} \\ $$$${p}=\mathrm{1}\Rightarrow\mathrm{2}{x}=\mathrm{2}{x}^{\mathrm{2}}…

x-2-3-x-3-1-3-1-x-3-1-3-1-3-x-5-6-x-3-1-3-1-2-x-3-1-3-1-3-5-

Question Number 197784 by cortano12 last updated on 28/Sep/23 $$\:\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{2}+\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\left(\mathrm{1}+\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\right)}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{x}+\mathrm{5}+\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\left(\mathrm{1}+\mathrm{2}\sqrt[{\mathrm{3}}]{\mathrm{x}−\mathrm{3}}\:\right)}\:=\:\mathrm{5} \\ $$ Answered by Frix last updated on 28/Sep/23 $$\mathrm{Let}\:{x}={t}^{\mathrm{3}} +\mathrm{3} \\ $$$$\sqrt[{\mathrm{3}}]{\left({t}+\mathrm{1}\right)^{\mathrm{3}} }+\sqrt[{\mathrm{3}}]{{t}^{\mathrm{3}} +\mathrm{12}{t}^{\mathrm{2}}…