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Question Number 176281 Answers: 0 Comments: 3
Question Number 173870 Answers: 0 Comments: 0
$$\sqrt{\mathrm{1}+\:\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)\left({n}+\mathrm{3}\right)\left({n}+\mathrm{4}\right)}\:\in\:\mathbb{N} \\ $$
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Question Number 173124 Answers: 2 Comments: 2
Question Number 172269 Answers: 0 Comments: 0
$${A}\:{rectangular}\:{picture}\:\mathrm{6}{cm}\:{by}\:\mathrm{8}{cm}\:{is} \\ $$$${enclosed}\:{by}\:{a}\:{frame}\:\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\:{wide}.\: \\ $$$${calculate}\:{the}\left[{area}\:{of}\:{the}\:{frame}.\right. \\ $$
Question Number 172266 Answers: 1 Comments: 0
Question Number 176848 Answers: 1 Comments: 0
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Question Number 170683 Answers: 0 Comments: 0
$${l}'{union}\:{d}'{un}\:{ferme}\:{et}\:{d}'{un}\:{borne}\:{est}-{il} \\ $$$${compacte}?\:{quand}\:{est}-{il}\:{de}\:{l}'{intersection}. \\ $$
Question Number 170552 Answers: 1 Comments: 1
Question Number 169088 Answers: 1 Comments: 0
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$${f}\left({x},{y},{z}\right)\:=\:\left(\mathrm{3}{x}^{\mathrm{2}} {y},{x}^{\mathrm{3}} +{y}^{\mathrm{3}} ,\:\mathrm{2}{z}\right) \\ $$$$\mathrm{prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{function}\:\mathrm{has}\:\mathrm{a}\:\mathrm{potential} \\ $$$$\mathrm{to}\:\mathrm{be}\:\mathrm{determined}. \\ $$
Question Number 168906 Answers: 1 Comments: 0
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Question Number 168233 Answers: 1 Comments: 0
$$\begin{cases}{{u}_{\mathrm{0}} \:=\:\mathrm{3}\::\:{u}_{\mathrm{1}} \:=\:\mathrm{4}}\\{{u}_{{n}+\mathrm{1}} \:=\:{u}_{{n}} \:+\:\mathrm{6}{u}_{{n}−\mathrm{1}} }\end{cases} \\ $$$${Express}\:{u}_{{n}} \:{in}\:{terms}\:{of}\:{n} \\ $$
Question Number 167520 Answers: 1 Comments: 0
$$\mathrm{Given}\:\mathrm{that}\:{f}\left({x}\right)\:=\:\int_{{x}} ^{\mathrm{2}{x}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{4}} }}{dt} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{state}\:\mathrm{its}\:\mathrm{domain} \\ $$$$\left(\mathrm{b}\right)\:\mathrm{is}\:{f}\left({x}\right)\:\mathrm{even}\:\mathrm{or}\:\mathrm{odd}? \\ $$
Question Number 165965 Answers: 2 Comments: 0
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Question Number 158751 Answers: 0 Comments: 0
$${Q}\:\mathrm{158528} \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathbb{P}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}}\right) \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}^{\mathrm{3}} }\right) \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{\left({n}+\mathrm{1}−\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}+{n}+\mathrm{1}+\mathrm{1}\right)}{\left({n}+\mathrm{1}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}−{n}−\mathrm{1}+\mathrm{1}\right)}\right\} \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{{n}}{{n}+\mathrm{2}}\right\}\bullet\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right\} \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left\{\frac{{k}}{{k}+\mathrm{2}}\right\}\bullet\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left\{\frac{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{3}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\right\} \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{3}}\bullet\frac{\mathrm{2}}{\mathrm{4}}\bullet\frac{\mathrm{3}}{\mathrm{5}}\bullet...\bullet\frac{{n}}{{n}+\mathrm{2}}\right\}×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{7}}{\mathrm{3}}\bullet\frac{\mathrm{13}}{\mathrm{7}}\bullet...\bullet\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right\} \\ $$$$=\mathrm{2}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right\}×\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}\right\} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\right\}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}+\frac{\mathrm{3}}{{n}}+\frac{\mathrm{3}}{{n}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{3}}{{n}}+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }}\right\}\:=\:\frac{\mathrm{2}}{\mathrm{3}}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...............\mathscr{L}{e}\:{puissant}............... \\ $$
Question Number 158740 Answers: 1 Comments: 1
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