Question Number 58769 by Mr X pcx last updated on 29/Apr/19 $${decompose}\:{the}\:{fractions}\:{inside}\:{C}\left({x}\right) \\ $$$$\left.\mathrm{1}\right)\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\left.\mathrm{2}\right)\:\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{5}} } \\ $$ Commented by maxmathsup…
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Question Number 189823 by Shrinava last updated on 22/Mar/23 $$\mathrm{If}:\:\:\:\mathrm{x}_{\mathrm{0}} \:=\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\mathrm{x}_{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:=\:\sqrt{\mathrm{x}_{\boldsymbol{\mathrm{n}}} ^{\mathrm{2}} \:+\:\mathrm{x}_{\boldsymbol{\mathrm{n}}} } \\ $$$$\mathrm{Find}:\:\:\:\Omega\:=\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\:\left(\mathrm{x}_{\boldsymbol{\mathrm{n}}} -\:\frac{\mathrm{n}}{\mathrm{2}}\:+\:\frac{\mathrm{1}}{\mathrm{8}}\:\underset{\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\boldsymbol{\mathrm{n}}−\mathrm{1}} {\sum}}\:\frac{\mathrm{1}}{\mathrm{x}_{\boldsymbol{\mathrm{k}}} }\right) \\…
Question Number 189803 by uchihayahia last updated on 22/Mar/23 $$ \\ $$$${what}'{s}\:{the}\:{minimum}\:{value}\:{of} \\ $$$${a}+\frac{\mathrm{1}}{{b}\left({a}−{b}\right)}\:{where}\:{a}>{b}>\mathrm{0}\:{a},{b}\in\mathbb{R} \\ $$$$ \\ $$ Answered by cortano12 last updated on 22/Mar/23…
Question Number 58716 by Jaiden2019 last updated on 28/Apr/19 $$\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$ Answered by Forkum Michael Choungong last updated on 28/Apr/19 $$=\:\frac{\mathrm{4}+\mathrm{3}}{\mathrm{12}}\:=\:\frac{\mathrm{7}}{\mathrm{12}} \\ $$$$ \\…
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Question Number 189757 by Shrinava last updated on 21/Mar/23 $$\mathrm{Find}:\:\:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\frac{\boldsymbol{\pi}}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\mid\:\mathrm{sin}\:\mathrm{4x}\:\mid}{\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{4x}}}\:\:\:=\:\:\:? \\ $$$$\mathrm{Find}:\:\:\:\:\:\underset{\boldsymbol{\mathrm{x}}\rightarrow\boldsymbol{\pi}} {\mathrm{lim}}\:\frac{\sqrt{\mathrm{2}}\:\mid\:\mathrm{sin}\:\mathrm{6x}\:\mid}{\:\sqrt{\mathrm{1}\:−\:\mathrm{cos}\:\mathrm{6x}}}\:\:\:=\:\:\:? \\ $$ Answered by Frix last updated on 21/Mar/23 $$\mathrm{Use}\:{t}=\mathrm{tan}\:\mathrm{2}{x}\:\mathrm{in}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{and}\:{t}=\mathrm{tan}\:\mathrm{3}{x}\:\mathrm{in}\:\mathrm{the}…
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Question Number 58682 by hhghg last updated on 27/Apr/19 $$\mathrm{3}\frac{\mathrm{1}}{\mathrm{5}}+\mathrm{2}\frac{\mathrm{1}}{\mathrm{15}} \\ $$ Answered by MJS last updated on 27/Apr/19 $$\mathrm{3}+\frac{\mathrm{1}}{\mathrm{5}}+\mathrm{2}+\frac{\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{1}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{1}^{×\mathrm{3}} }{\mathrm{5}^{×\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{15}}= \\ $$$$=\mathrm{5}+\frac{\mathrm{3}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{3}+\mathrm{1}}{\mathrm{15}}=\mathrm{5}+\frac{\mathrm{4}}{\mathrm{15}}=\mathrm{5}\frac{\mathrm{4}}{\mathrm{15}} \\…