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Question Number 163467 Answers: 1 Comments: 0
Question Number 163463 Answers: 1 Comments: 0
$${nature}\:{de}\:{la}\:{serie} \\ $$$$\underset{{n}=\mathrm{1}} {\sum}\left(\frac{\mathrm{1}}{{n}+\mathrm{1}}+\frac{\mathrm{1}}{{n}}\right) \\ $$
Question Number 163457 Answers: 0 Comments: 1
$$\mathrm{let}\:\:\boldsymbol{\mathrm{a}}>\mathrm{0}\:\:\mathrm{and}\:\:\boldsymbol{\lambda}>\mathrm{0}\:\:\mathrm{fixed} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\:\left(\mathrm{0};\infty\right)\:\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{2a}^{\mathrm{2}} \mathrm{cos}\left(\frac{\mathrm{x}}{\mathrm{2}\lambda}\:-\:\frac{\mathrm{2}\lambda}{\mathrm{x}}\right)\:=\:\mathrm{a}^{\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\lambda}}} \:\:+\:\:\mathrm{a}^{\frac{\mathrm{4}\boldsymbol{\lambda}}{\boldsymbol{\mathrm{x}}}} \\ $$
Question Number 163452 Answers: 1 Comments: 1
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{3}^{\boldsymbol{\mathrm{x}}\:\sqrt{\boldsymbol{\mathrm{x}}}} \:\:+\:\:\mathrm{3}^{\mathrm{1}\:+\:\frac{\mathrm{1}}{\:\sqrt{\boldsymbol{\mathrm{x}}}}} \:\:=\:\mathrm{12} \\ $$
Question Number 163451 Answers: 0 Comments: 1
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\mathrm{1}\:-\:\mathrm{x}}\:=\:\mathrm{1}\:-\:\mathrm{2x}^{\mathrm{2}} \:+\:\mathrm{2x}\:\sqrt{\mathrm{1}\:-\:\mathrm{x}^{\mathrm{2}} } \\ $$
Question Number 163443 Answers: 0 Comments: 7
$$\mathrm{p}\:\leqslant\:\mathrm{n}\: \\ $$$$\mathrm{find}\:\:\frac{\mathrm{A}_{\mathrm{n}} ^{\mathrm{p}} }{\mathrm{A}_{\mathrm{n}β\mathrm{1}} ^{\mathrm{p}} }. \\ $$
Question Number 163441 Answers: 0 Comments: 0
$$\int\frac{{x}+\mathrm{3}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx}=\int\frac{{x}+\mathrm{1}+\mathrm{3}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx}=\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{3}} }{dx}+\mathrm{3}\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{4}} }{dx} \\ $$
Question Number 163439 Answers: 0 Comments: 0
Question Number 163437 Answers: 1 Comments: 2
Question Number 163434 Answers: 0 Comments: 0
Question Number 163433 Answers: 0 Comments: 0
Question Number 163432 Answers: 1 Comments: 0
Question Number 163430 Answers: 1 Comments: 0
Question Number 163428 Answers: 0 Comments: 2
$${jusgifier}\:{la}\:{convergence}\:{de}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left(\mathrm{1}β{x}^{\mathrm{2}} \right){dx} \\ $$
Question Number 163414 Answers: 1 Comments: 0
$$\int\frac{\boldsymbol{{dx}}}{\:\sqrt{\boldsymbol{{cosx}}\:\boldsymbol{{sin}}^{\mathrm{3}} \boldsymbol{{x}}}+\sqrt{\boldsymbol{{sinx}}\:\boldsymbol{{cos}}^{\mathrm{3}} \boldsymbol{{x}}}} \\ $$
Question Number 163413 Answers: 0 Comments: 3
Question Number 163408 Answers: 1 Comments: 0
$$\sqrt[{\mathrm{3}}]{\mathrm{1}+\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}+\sqrt[{\mathrm{3}}]{\mathrm{1}β\frac{\mathrm{2}}{\mathrm{9}}\sqrt{\mathrm{21}}}=? \\ $$
Question Number 163402 Answers: 1 Comments: 0
$$\:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \left(\boldsymbol{{x}}β\mathrm{1}\right)^{\mathrm{10}} \left(\boldsymbol{{x}}β\mathrm{3}\right)^{\mathrm{3}} \boldsymbol{{dx}} \\ $$
Question Number 163411 Answers: 0 Comments: 0
Question Number 163400 Answers: 2 Comments: 0
$$ \\ $$$$\:\:\:\:\:{prove}\: \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\infty} {cot}^{\:β\mathrm{1}} \left(\mathrm{1}+{x}^{\:\mathrm{2}} \right)=\left(\sqrt{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}β\frac{\mathrm{1}}{\mathrm{2}}}\:\right)\:\:\pi \\ $$$$ \\ $$
Question Number 163397 Answers: 0 Comments: 5
Question Number 163396 Answers: 0 Comments: 0
$$\mathrm{if}\:\:\boldsymbol{\mathrm{a}}\:\:\mathrm{and}\:\:\boldsymbol{\mathrm{b}}\:\:\mathrm{are}\:\mathrm{positive}\:\mathrm{numbers} \\ $$$$\mathrm{then}\:\:\Sigma\:\left(\mathrm{an}\:+\:\mathrm{b}\right)^{-\boldsymbol{\mathrm{p}}} \:\:\mathrm{converges}\:\mathrm{if}\:\:\boldsymbol{\mathrm{p}}>\mathrm{1} \\ $$$$\mathrm{and}\:\mathrm{diverges}\:\mathrm{if}\:\:\boldsymbol{\mathrm{p}}\leqslant\mathrm{1} \\ $$
Question Number 163393 Answers: 1 Comments: 0
Question Number 163386 Answers: 1 Comments: 0
$$\underset{{x}\rightarrowβ\infty} {\mathrm{lim}}\sqrt[{\mathrm{3}}]{\mathrm{1}β{x}^{\mathrm{3}} }β\sqrt[{\mathrm{4}}]{{x}^{\mathrm{4}} β\mathrm{1}} \\ $$
Question Number 163385 Answers: 1 Comments: 0
Question Number 163384 Answers: 2 Comments: 0
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