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Question Number 158812 Answers: 0 Comments: 0

Question Number 158805 Answers: 0 Comments: 0

(1)F(x)= x^3 [ x ] ⇒ { ((F ′(0)=?)),((F ′(1)=?)) :} (2) F(x)= [ x ]−∣x∣ ⇒F ′(−(5/2))=? where [ ] : floor function ∣ ∣ absolute function

$$\left(\mathrm{1}\right){F}\left({x}\right)=\:{x}^{\mathrm{3}} \:\left[\:{x}\:\right]\:\Rightarrow\begin{cases}{{F}\:'\left(\mathrm{0}\right)=?}\\{{F}\:'\left(\mathrm{1}\right)=?}\end{cases} \\ $$$$\:\left(\mathrm{2}\right)\:{F}\left({x}\right)=\:\left[\:{x}\:\right]−\mid{x}\mid\:\Rightarrow{F}\:'\left(−\frac{\mathrm{5}}{\mathrm{2}}\right)=? \\ $$$$\:{where}\:\left[\:\right]\::\:{floor}\:{function} \\ $$$$\:\mid\:\mid\:{absolute}\:{function}\: \\ $$

Question Number 158803 Answers: 1 Comments: 1

Question Number 158794 Answers: 2 Comments: 0

montrer que 7 divise 2222^(5555) +5555^(2222)

$${montrer}\:{que}\:\mathrm{7}\:{divise} \\ $$$$\mathrm{2222}^{\mathrm{5555}} +\mathrm{5555}^{\mathrm{2222}} \\ $$$$ \\ $$

Question Number 158774 Answers: 1 Comments: 0

Question Number 158775 Answers: 1 Comments: 0

Find: ∫_( 0) ^( ∞) (1/((x^2 + x + 1)∙(1 + ax))) dx ; a>0 Answer: ((-π(√3)∙(a-2)+9a∙ln(a))/(9∙(a^2 -a+1)))

$$\mathrm{Find}: \\ $$$$\underset{\:\mathrm{0}} {\overset{\:\infty} {\int}}\:\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{1}\right)\centerdot\left(\mathrm{1}\:+\:\mathrm{ax}\right)}\:\mathrm{dx}\:\:;\:\:\mathrm{a}>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Answer}: \\ $$$$\frac{-\pi\sqrt{\mathrm{3}}\centerdot\left(\mathrm{a}-\mathrm{2}\right)+\mathrm{9a}\centerdot\mathrm{ln}\left(\mathrm{a}\right)}{\mathrm{9}\centerdot\left(\mathrm{a}^{\mathrm{2}} -\mathrm{a}+\mathrm{1}\right)} \\ $$

Question Number 158768 Answers: 1 Comments: 0

evaluate ∫2x(√(4x−5)) dx

$$ \\ $$$$\mathrm{evaluate} \\ $$$$\int\mathrm{2x}\sqrt{\mathrm{4x}−\mathrm{5}}\:\mathrm{dx} \\ $$

Question Number 158761 Answers: 0 Comments: 0

Question Number 158760 Answers: 1 Comments: 2

f(x)=[sgn(x^2 −1)+sgn(sin πx)] faind lim_(x→1) f(x)=?

$${f}\left({x}\right)=\left[{sgn}\left({x}^{\mathrm{2}} −\mathrm{1}\right)+{sgn}\left(\mathrm{sin}\:\pi{x}\right)\right] \\ $$$${faind}\:\:\:\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}{f}\left({x}\right)=? \\ $$

Question Number 158759 Answers: 1 Comments: 0

Prove that: lim_(n→∞) ((Σ_(k=0) ^(2n) (-1)^k ∙ ((4n + 1)/(4n - 2k + 1)) (((2n)),(( k)) )))^(1/n) = 1

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\underset{\boldsymbol{\mathrm{n}}\rightarrow\infty} {\mathrm{lim}}\sqrt[{\boldsymbol{\mathrm{n}}}]{\underset{\boldsymbol{\mathrm{k}}=\mathrm{0}} {\overset{\mathrm{2}\boldsymbol{\mathrm{n}}} {\sum}}\left(-\mathrm{1}\right)^{\boldsymbol{\mathrm{k}}} \:\centerdot\:\frac{\mathrm{4n}\:+\:\mathrm{1}}{\mathrm{4n}\:-\:\mathrm{2k}\:+\:\mathrm{1}}\begin{pmatrix}{\mathrm{2n}}\\{\:\mathrm{k}}\end{pmatrix}}\:=\:\mathrm{1} \\ $$$$ \\ $$

Question Number 158751 Answers: 0 Comments: 0

Q 158528 P=Π_(n=1) ^∞ ((((n+1)^3 −1)/((n+1)^3 +1))) ⇒ P = Π_(n=1) ^∞ ((((n+1)^3 −1^3 )/((n+1)^3 +1^3 ))) ⇒ P = Π_(n=1) ^∞ {(((n+1−1)(n^2 +2n+1+n+1+1))/((n+1+1)(n^2 +2n+1−n−1+1)))} ⇒ P = Π_(n=1) ^∞ {(n/(n+2))}•Π_(n=1) ^∞ {((n^2 +3n+3)/(n^2 +n+1))} = lim_(n→∞) Π_(k=1) ^n {(k/(k+2))}•lim_(n→∞) Π_(k=1) ^n {((k^2 +3k+3)/(k^2 +k+1))} = lim_(n→∞) {(1/3)•(2/4)•(3/5)•...•(n/(n+2))}×lim_(n→∞) {(7/3)•((13)/7)•...•((n^2 +3n+3)/(n^2 +n+1))} =2lim_(n→∞) {(1/((n+1)(n+2)))}×(1/3)lim_(n→∞) {n^2 +3n+3} =(2/3)lim_(n→∞) {((n^2 +3n+3)/(n^2 +3n+2))} = (2/3)lim_(n→∞) {((1+(3/n)+(3/n^2 ))/(1+(3/n)+(2/n^2 )))} = (2/3). P = Π_(n=1) ^∞ ((((n+1)^3 −1)/((n+1)^3 +1))) = (2/3).. ...............Le puissant...............

$${Q}\:\mathrm{158528} \\ $$$$ \\ $$$$\:\:\:\:\:\:\mathbb{P}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}}\right) \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}^{\mathrm{3}} }{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}^{\mathrm{3}} }\right) \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{\left({n}+\mathrm{1}−\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}+{n}+\mathrm{1}+\mathrm{1}\right)}{\left({n}+\mathrm{1}+\mathrm{1}\right)\left({n}^{\mathrm{2}} +\mathrm{2}{n}+\mathrm{1}−{n}−\mathrm{1}+\mathrm{1}\right)}\right\} \\ $$$$\Rightarrow\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{{n}}{{n}+\mathrm{2}}\right\}\bullet\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left\{\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right\} \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left\{\frac{{k}}{{k}+\mathrm{2}}\right\}\bullet\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left\{\frac{{k}^{\mathrm{2}} +\mathrm{3}{k}+\mathrm{3}}{{k}^{\mathrm{2}} +{k}+\mathrm{1}}\right\} \\ $$$$=\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\mathrm{3}}\bullet\frac{\mathrm{2}}{\mathrm{4}}\bullet\frac{\mathrm{3}}{\mathrm{5}}\bullet...\bullet\frac{{n}}{{n}+\mathrm{2}}\right\}×\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{7}}{\mathrm{3}}\bullet\frac{\mathrm{13}}{\mathrm{7}}\bullet...\bullet\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +{n}+\mathrm{1}}\right\} \\ $$$$=\mathrm{2}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}\right\}×\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}\right\} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{3}}{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}\right\}\:=\:\frac{\mathrm{2}}{\mathrm{3}}\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\frac{\mathrm{1}+\frac{\mathrm{3}}{{n}}+\frac{\mathrm{3}}{{n}^{\mathrm{2}} }}{\mathrm{1}+\frac{\mathrm{3}}{{n}}+\frac{\mathrm{2}}{{n}^{\mathrm{2}} }}\right\}\:=\:\frac{\mathrm{2}}{\mathrm{3}}. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathbb{P}\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\prod}}\left(\frac{\left({n}+\mathrm{1}\right)^{\mathrm{3}} −\mathrm{1}}{\left({n}+\mathrm{1}\right)^{\mathrm{3}} +\mathrm{1}}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}.. \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:...............\mathscr{L}{e}\:{puissant}............... \\ $$

Question Number 158749 Answers: 2 Comments: 1

Question Number 158742 Answers: 1 Comments: 1

Question Number 158740 Answers: 1 Comments: 1

Question Number 158735 Answers: 0 Comments: 2

Question Number 158731 Answers: 0 Comments: 0

Question Number 158708 Answers: 2 Comments: 1

Question Number 158707 Answers: 0 Comments: 0

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Question Number 158724 Answers: 2 Comments: 0

let a>b>c>0 solve in R { ((ax + by + cz = a)),((bx + cy + az = b)),((cx + ay + bz = c)) :}

$$\mathrm{let}\:\:\mathrm{a}>\mathrm{b}>\mathrm{c}>\mathrm{0}\:\:\mathrm{solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\begin{cases}{\mathrm{ax}\:+\:\mathrm{by}\:+\:\mathrm{cz}\:=\:\mathrm{a}}\\{\mathrm{bx}\:+\:\mathrm{cy}\:+\:\mathrm{az}\:=\:\mathrm{b}}\\{\mathrm{cx}\:+\:\mathrm{ay}\:+\:\mathrm{bz}\:=\:\mathrm{c}}\end{cases} \\ $$$$ \\ $$

Question Number 158700 Answers: 0 Comments: 0

Question Number 158699 Answers: 0 Comments: 3

(√((log_3 3(√(3x))+log_x 3(√(3x)))log_3 x^3 ))+(√((((log_3 3(√x))/3)+((log_x 3(√x))/3))log_3 x^3 ))=2 please i need help. Ive been trying but still not getting answer.

$$\sqrt{\left(\mathrm{log}_{\mathrm{3}} \mathrm{3}\sqrt{\mathrm{3x}}+\mathrm{log}_{\mathrm{x}} \mathrm{3}\sqrt{\mathrm{3x}}\right)\mathrm{log}_{\mathrm{3}} \mathrm{x}^{\mathrm{3}} }+\sqrt{\left(\frac{\mathrm{log}_{\mathrm{3}} \mathrm{3}\sqrt{\mathrm{x}}}{\mathrm{3}}+\frac{\mathrm{log}_{\mathrm{x}} \mathrm{3}\sqrt{\mathrm{x}}}{\mathrm{3}}\right)\mathrm{log}_{\mathrm{3}} \mathrm{x}^{\mathrm{3}} }=\mathrm{2} \\ $$$$\mathrm{please}\:\mathrm{i}\:\mathrm{need}\:\mathrm{help}. \\ $$$$\mathrm{Ive}\:\mathrm{been}\:\mathrm{trying}\:\mathrm{but}\:\mathrm{still}\:\mathrm{not}\:\mathrm{getting}\: \\ $$$$\mathrm{answer}. \\ $$

Question Number 158698 Answers: 0 Comments: 0

(x+1)^2 (x^2 −4c^2 )=4c^4

$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} \left({x}^{\mathrm{2}} −\mathrm{4}{c}^{\mathrm{2}} \right)=\mathrm{4}{c}^{\mathrm{4}} \\ $$

Question Number 158697 Answers: 1 Comments: 0

∫_0 ^1 ln(ln(1−x))dx=?

$$\int_{\mathrm{0}} ^{\mathrm{1}} {ln}\left({ln}\left(\mathrm{1}−{x}\right)\right){dx}=? \\ $$

Question Number 158696 Answers: 0 Comments: 0

find the partial differention equation if u = (g o f ) (x+y) help me sir

$${find}\:{the}\:{partial}\:{differention}\:{equation}\:{if}\: \\ $$$${u}\:=\:\left({g}\:{o}\:{f}\:\right)\:\left({x}+{y}\right) \\ $$$$ \\ $$$${help}\:{me}\:{sir} \\ $$

Question Number 158691 Answers: 1 Comments: 0

Σ_(n=1) ^∞ (( (−1)^( n) H_( 2n) )/(2n)) =?

$$ \\ $$$$\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\:{n}} \:\mathcal{H}_{\:\mathrm{2}{n}} }{\mathrm{2}{n}}\:=? \\ $$

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