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Question Number 144143 Answers: 1 Comments: 0
$$\int_{\frac{\mathrm{1}}{{a}}} ^{{a}} \frac{{arctg}\left({x}\right)}{{x}}{dx}=??? \\ $$
Question Number 144142 Answers: 1 Comments: 0
$$\:\int\:\frac{\sqrt{\mathrm{cos}\:\mathrm{x}+\sqrt{\mathrm{cos}\:\mathrm{x}+\sqrt{\mathrm{cos}\:\mathrm{x}+\sqrt{\mathrm{cos}\:\mathrm{x}+\sqrt{...}}}}}}{\mathrm{sin}\:\mathrm{x}}\:\mathrm{dx} \\ $$
Question Number 144139 Answers: 2 Comments: 0
Question Number 144136 Answers: 0 Comments: 1
Question Number 144120 Answers: 2 Comments: 0
$${log}_{\mathrm{2}} \mathrm{3}\:=\:{x}\:,\:{log}_{\mathrm{3}} \mathrm{5}\:=\:{y}\:,\:{lg}\mathrm{6}\:=\:? \\ $$
Question Number 144118 Answers: 0 Comments: 0
$$\mathrm{Let}\:{a},{b},{c}\:>\:\mathrm{0}\:\mathrm{and}\:{a}+{b}+{c}\:=\:\mathrm{3}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\sqrt{{ab}}+\mathrm{1}}{\:\sqrt{{ab}}+\sqrt{{c}}}+\frac{\sqrt{{bc}}+\mathrm{1}}{\:\sqrt{{bc}}+\sqrt{{a}}}+\frac{\sqrt{{ca}}+\mathrm{1}}{\:\sqrt{{ca}}+\sqrt{{b}}}\:\geqslant\:\sqrt{{a}}+\sqrt{{b}}+\sqrt{{c}}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$
Question Number 144116 Answers: 0 Comments: 0
Question Number 144109 Answers: 3 Comments: 1
$$\boldsymbol{\mathrm{hi}},\:\boldsymbol{\mathrm{everybody}}\:! \\ $$$$\mathrm{1}.\:\boldsymbol{\mathrm{calculate}}\::\:\boldsymbol{\mathrm{I}}\:=\int_{\frac{\boldsymbol{\pi}}{\mathrm{6}}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{3}}} \boldsymbol{{ln}}\left(\boldsymbol{{tan}}\:\boldsymbol{{x}}\right)\boldsymbol{{dx}}. \\ $$$$\mathrm{2}.\:\boldsymbol{\mathrm{calculate}}\:\::\:\underset{\boldsymbol{{x}}\:\rightarrow\:\boldsymbol{{e}}} {\boldsymbol{{lim}}}\:\frac{\boldsymbol{{x}}\sqrt{\mathrm{1}−\boldsymbol{{ln}}\:\boldsymbol{{x}}}}{\boldsymbol{{x}}−\boldsymbol{{e}}}\:. \\ $$
Question Number 144107 Answers: 1 Comments: 0
$$\mathrm{A}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\frac{\mathrm{1}+\sqrt[{\mathrm{7}}]{\mathrm{2}}+\sqrt[{\mathrm{7}}]{\mathrm{3}}+\sqrt[{\mathrm{7}}]{\mathrm{4}}+.....+\sqrt[{\mathrm{7}}]{\mathrm{n}}}{\:\sqrt[{\mathrm{7}}]{\mathrm{n}^{\mathrm{9}} }}\:=? \\ $$
Question Number 144106 Answers: 1 Comments: 1
$$\underset{{x}\rightarrow\infty} {{lim}}\frac{\mathrm{4}^{{x}-\mathrm{2}} \:+\:\mathrm{3}^{{x}} \:+\:\mathrm{2}^{{x}} }{\mathrm{4}^{{x}-\mathrm{1}} \:+\:\mathrm{3}^{{x}+\mathrm{1}} }\:=\:? \\ $$
Question Number 144095 Answers: 1 Comments: 0
$${if}\:\:{x}>\mathrm{0}\:,\:{r}={pq}\:,\:\mathrm{1}\leqslant{p}\leqslant{q}\:\:{then}: \\ $$$$\mathrm{1}+{rx}\:\leqslant\:\left(\mathrm{1}+{qx}\right)^{\boldsymbol{{p}}} \:\leqslant\:\left(\mathrm{1}+{px}\right)^{\boldsymbol{{q}}} \:\leqslant\:\left(\mathrm{1}+{x}\right)^{\boldsymbol{{r}}} \\ $$
Question Number 144094 Answers: 1 Comments: 1
Question Number 144083 Answers: 1 Comments: 0
$${if}\:\:{a};{b};{c}>\mathrm{0}\:\:{then}: \\ $$$$\frac{\mathrm{9}{abc}}{{a}^{\mathrm{3}} \:+\:{b}^{\mathrm{3}} \:+\:{c}^{\mathrm{3}} }\:+\:\frac{{a}^{\mathrm{2}} }{{bc}}\:+\:\frac{{b}^{\mathrm{2}} }{{ca}}\:+\:\frac{{c}^{\mathrm{2}} }{{ab}}\:\geqslant\:\mathrm{6} \\ $$
Question Number 144074 Answers: 1 Comments: 0
$$\:\mathrm{The}\:\mathrm{parallelogram}\:\mathrm{ABCD}\:\mathrm{has} \\ $$$$\mid\mid\mathrm{AB}\mid\mid\:=\mathrm{6},\:\mid\mid\mathrm{AC}\mid\mid=\mathrm{7}\:\&\:\mathrm{d}\left(\mathrm{D},\mathrm{AC}\right)=\mathrm{2} \\ $$$$\mathrm{Find}\:\mathrm{d}\left(\mathrm{D},\mathrm{AB}\right). \\ $$
Question Number 144072 Answers: 2 Comments: 0
$$\mathrm{Use}\:\mathrm{the}\:\mathrm{reduction}\:\mathrm{formula}\:\mathrm{to} \\ $$$$\mathrm{rewrite}\:−\mathrm{3}\:\mathrm{sin}\:\mathrm{x}\:−\mathrm{3}\:\mathrm{cos}\:\mathrm{x}\:\mathrm{in}\:\mathrm{the} \\ $$$$\mathrm{form}\:\mathrm{K}\:\mathrm{sin}\:\left(\mathrm{x}+\alpha\right)\:. \\ $$
Question Number 144069 Answers: 0 Comments: 0
$$\mathrm{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{2304cosx}}{\left(\mathrm{cos4x}−\mathrm{8cos2x}+\mathrm{15}\right)^{\mathrm{2}} }\mathrm{dx}=? \\ $$
Question Number 144068 Answers: 1 Comments: 0
$$\mathrm{S}_{\mathrm{2019}} =\mathrm{1}+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:...+\:\frac{\mathrm{1}}{\mathrm{2019}^{\mathrm{2}} }\:=? \\ $$
Question Number 144067 Answers: 1 Comments: 0
$$\mathrm{L}=\underset{\mathrm{n}\rightarrow+\propto} {\mathrm{lim}}\left(\frac{\mathrm{1}^{\mathrm{2}} }{\mathrm{n}^{\mathrm{3}} +\mathrm{1}^{\mathrm{3}} }\:+\:\frac{\mathrm{2}^{\mathrm{3}} }{\mathrm{n}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} }\:+\:...+\:\frac{\mathrm{n}^{\mathrm{3}} }{\mathrm{n}^{\mathrm{3}} +\mathrm{n}^{\mathrm{3}} }\right)=? \\ $$
Question Number 144064 Answers: 2 Comments: 0
Question Number 144062 Answers: 0 Comments: 0
Question Number 144061 Answers: 1 Comments: 0
Question Number 144060 Answers: 2 Comments: 1
$${p}\left({x}\right)=\left({x}−\mathrm{1}\right)^{\mathrm{2}} {Q}\left({x}\right)+\mathrm{3}{x}+\mathrm{8} \\ $$$${p}\left({x}\right)=\left({x}−\mathrm{2}\right){Q}\left({x}\right)+{R} \\ $$$${R}=? \\ $$
Question Number 144059 Answers: 1 Comments: 0
Question Number 144057 Answers: 2 Comments: 0
$${En}\:{utilisant}\:{la}\:{transforme}\:{de}\:{laplace} \\ $$$${Calculer} \\ $$$$\int_{\mathrm{0}} ^{+\infty} \frac{{tsin}\left({xt}\right)}{{a}^{\mathrm{2}} +{t}^{\mathrm{2}} }{dt}\:\:\:\forall{a},{x}\in\mathbb{R}^{\ast} \\ $$
Question Number 144053 Answers: 1 Comments: 0
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:........\:{Calculus}........ \\ $$$$\:\:\:\:\:\:\:\Omega:={lim}\frac{\mathrm{1}}{\pi}\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{sin}\left({kx}\right)}{\:\sqrt{\mathrm{2}^{{k}} }}\right)^{\mathrm{2}} {dx}=? \\ $$$$ \\ $$
Question Number 144052 Answers: 1 Comments: 0
$$\mathrm{Evaluate}\: \\ $$$$\:\int\:\frac{\sqrt{{x}}}{\mathrm{sinh}\:{x}}\:{dx} \\ $$
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