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Question Number 162234 Answers: 0 Comments: 0

Σ_(p=0) ^n (−1)^p C_p ^(1/2) (−1)^(n−p) C_(n−p) ^(1/2) = ???? Please Help me(Aidez moi)

$$ \\ $$$$\:\:\:\:\:\:\underset{{p}=\mathrm{0}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\right)^{{p}} {C}_{{p}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \left(−\mathrm{1}\right)^{{n}−{p}} {C}_{{n}−{p}} ^{\frac{\mathrm{1}}{\mathrm{2}}} \:=\:\:\:???? \\ $$$$\:\:\:\:\:\:\:{Please}\:{Help}\:{me}\left({Aidez}\:{moi}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 162229 Answers: 1 Comments: 0

Question Number 162287 Answers: 0 Comments: 4

lim_(n→∞) ∫_0 ^1 ((x^n +(1−x)^n ))^(1/n) dx=?

$$\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt[{\mathrm{n}}]{\mathrm{x}^{\mathrm{n}} +\left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{n}} }\mathrm{dx}=? \\ $$

Question Number 162219 Answers: 2 Comments: 0

Ω=∫_0 ^1 ((log(1+x^7 ))/(1+x^7 ))dx=?

$$\Omega=\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{log}\left(\mathrm{1}+{x}^{\mathrm{7}} \right)}{\mathrm{1}+{x}^{\mathrm{7}} }{dx}=? \\ $$

Question Number 162209 Answers: 1 Comments: 2

Question Number 162190 Answers: 1 Comments: 0

⌊ ((3^2 +1)/(3^2 −1)) + ((3^3 +1)/(3^3 −1)) + ((3^4 +1)/(3^4 −1)) + …+ ((3^(2017) +1)/(3^(2017) −1)) ⌋ = ?

$$\lfloor\:\frac{\mathrm{3}^{\mathrm{2}} +\mathrm{1}}{\mathrm{3}^{\mathrm{2}} −\mathrm{1}}\:+\:\frac{\mathrm{3}^{\mathrm{3}} +\mathrm{1}}{\mathrm{3}^{\mathrm{3}} −\mathrm{1}}\:+\:\frac{\mathrm{3}^{\mathrm{4}} +\mathrm{1}}{\mathrm{3}^{\mathrm{4}} −\mathrm{1}}\:+\:\ldots+\:\frac{\mathrm{3}^{\mathrm{2017}} +\mathrm{1}}{\mathrm{3}^{\mathrm{2017}} −\mathrm{1}}\:\rfloor\:=\:\:? \\ $$

Question Number 162189 Answers: 1 Comments: 2

show the converge^ nce and calculate ∫_(−1) ^1 (√((1−t)/(1+t))) dt

$${show}\:{the}\:{converge}^{} {nce}\:{and}\:{calculate} \\ $$$$\int_{−\mathrm{1}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−{t}}{\mathrm{1}+{t}}}\:{dt} \\ $$

Question Number 162187 Answers: 0 Comments: 1

x^y =9 ((32y))^(1/y) =2x^2 (x,y)=? sulotion =?

$${x}^{{y}} =\mathrm{9} \\ $$$$\sqrt[{{y}}]{\mathrm{32}{y}}=\mathrm{2}{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\left({x},{y}\right)=? \\ $$$${sulotion}\:=? \\ $$$$ \\ $$

Question Number 162183 Answers: 1 Comments: 0

Question Number 162182 Answers: 1 Comments: 0

lim_(x→∞) (((x−(√(x^2 −3x+2)) )^(2022) +(x−(√(x^2 −5)) )^(2022) )/x^(2022) ) = ?

$$\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+\mathrm{2}}\:\right)^{\mathrm{2022}} +\left(\mathrm{x}−\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{5}}\:\right)^{\mathrm{2022}} }{\mathrm{x}^{\mathrm{2022}} }\:=\:? \\ $$

Question Number 162181 Answers: 1 Comments: 1

Solve the system of equations {: ((x^4 −2x+y=y^4 )),(((x^2 −y^2 )^3 = 3)) }

$$\:\:\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\: \\ $$$$\:\:\:\:\:\left.\begin{matrix}{\mathrm{x}^{\mathrm{4}} −\mathrm{2x}+\mathrm{y}=\mathrm{y}^{\mathrm{4}} }\\{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} \right)^{\mathrm{3}} \:=\:\mathrm{3}}\end{matrix}\right\}\: \\ $$

Question Number 162177 Answers: 1 Comments: 0

𝛗 = ∫_0 ^( 1) (( ln^( 2) ( x ). Li_( 2) (x ))/x^ ) dx =?

$$ \\ $$$$\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}^{\:\mathrm{2}} \left(\:{x}\:\right).\:\mathrm{Li}_{\:\mathrm{2}} \:\left({x}\:\right)}{{x}^{\:} }\:{dx}\:=? \\ $$

Question Number 162174 Answers: 0 Comments: 1

x^( 2) − 4x −1=0 α , β are roots α^( 3) + 17β +5 =? −−−solution−−− α is root ⇒ α^( 2) −4α −1=0 ⇒ α^( 2) = 4α +1 ✓ α^( 3) + 17β +5 = α . α^( 2) +17β +5 = α ( 4α +1 )+ 17β +5 = 4α^( 2) + α + 17β +5 = 4 (4α +1 )+α +17β +5=17(α+β)+9 = 17S +9= 17 (4 )+9=77

$$ \\ $$$$\:\:\:\:{x}^{\:\mathrm{2}} −\:\mathrm{4}{x}\:−\mathrm{1}=\mathrm{0}\:\: \\ $$$$\:\:\:\:\:\alpha\:,\:\beta\:\:{are}\:{roots}\: \\ $$$$\:\:\:\:\:\alpha^{\:\mathrm{3}} \:+\:\mathrm{17}\beta\:+\mathrm{5}\:=? \\ $$$$\:\:−−−{solution}−−− \\ $$$$\:\:\:\alpha\:\:\:{is}\:{root}\:\:\:\Rightarrow\:\alpha^{\:\mathrm{2}} −\mathrm{4}\alpha\:−\mathrm{1}=\mathrm{0} \\ $$$$\:\:\:\:\:\Rightarrow\:\alpha^{\:\mathrm{2}} =\:\mathrm{4}\alpha\:+\mathrm{1}\:\:\checkmark \\ $$$$\:\:\:\:\:\:\alpha^{\:\mathrm{3}} +\:\mathrm{17}\beta\:+\mathrm{5}\:=\:\alpha\:.\:\alpha^{\:\mathrm{2}} +\mathrm{17}\beta\:+\mathrm{5} \\ $$$$\:\:=\:\alpha\:\left(\:\mathrm{4}\alpha\:+\mathrm{1}\:\right)+\:\mathrm{17}\beta\:+\mathrm{5} \\ $$$$\:\:=\:\mathrm{4}\alpha^{\:\mathrm{2}} +\:\alpha\:+\:\mathrm{17}\beta\:+\mathrm{5} \\ $$$$\:\:=\:\mathrm{4}\:\left(\mathrm{4}\alpha\:+\mathrm{1}\:\right)+\alpha\:+\mathrm{17}\beta\:+\mathrm{5}=\mathrm{17}\left(\alpha+\beta\right)+\mathrm{9} \\ $$$$\:\:=\:\mathrm{17S}\:+\mathrm{9}=\:\mathrm{17}\:\left(\mathrm{4}\:\right)+\mathrm{9}=\mathrm{77} \\ $$$$ \\ $$

Question Number 162171 Answers: 0 Comments: 0

Question Number 162169 Answers: 2 Comments: 0

determinant ((( determinant ((( x+y+x^2 y^2 =586_(x=?,y=? ) ^(x,y∈Z ) )))_ ^ _() ^(•) )))

$$ \\ $$$$\:\:\:\:\:\:\:\begin{array}{|c|}{\overset{\bullet} {\:\:\:\:\:\begin{array}{|c|}{\:\:\:\underset{{x}=?,{y}=?\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {\overset{{x},{y}\in\mathbb{Z}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:} {{x}+{y}+{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{586}}}\:\:}\\\hline\end{array}_{} ^{} }\:\:\:\:}\\\hline\end{array} \\ $$$$ \\ $$

Question Number 162168 Answers: 0 Comments: 0

Solve the integro−differential equation: i(t) + 4(di/dt) + ∫i(t)dt = 2 cos (3t+ 60°) where i(t) is a sinulsodial current.

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{integro}−\mathrm{differential} \\ $$$$\mathrm{equation}: \\ $$$$\:{i}\left({t}\right)\:+\:\mathrm{4}\frac{{di}}{{dt}}\:+\:\int{i}\left({t}\right){dt}\:=\:\mathrm{2}\:\mathrm{cos}\:\left(\mathrm{3}{t}+\:\mathrm{60}°\right) \\ $$$$\mathrm{where}\:{i}\left({t}\right)\:\mathrm{is}\:\mathrm{a}\:\mathrm{sinulsodial}\:\mathrm{current}. \\ $$

Question Number 162151 Answers: 1 Comments: 0

Question Number 162116 Answers: 2 Comments: 0

A function f is such that f : R → R where f(xy+1) = f(x)∙f(y)−f(y)−x+2 , ∀x,y ∈ R . Find value of 10∙f(2017)+f(0) .

$${A}\:{function}\:\:{f}\:\:\:{is}\:\:{such}\:\:{that}\:\:{f}\::\:\mathbb{R}\:\rightarrow\:\mathbb{R}\:\:{where} \\ $$$$\:\:\:{f}\left({xy}+\mathrm{1}\right)\:=\:{f}\left({x}\right)\centerdot{f}\left({y}\right)−{f}\left({y}\right)−{x}+\mathrm{2}\:\:,\:\:\forall{x},{y}\:\in\:\mathbb{R}\:. \\ $$$${Find}\:\:{value}\:\:{of}\:\:\mathrm{10}\centerdot{f}\left(\mathrm{2017}\right)+{f}\left(\mathrm{0}\right)\:. \\ $$

Question Number 162112 Answers: 1 Comments: 0

∫(( cos(x))/((1−cos(x))^2 ))dx=?

$$\int\frac{\:\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)}{\left(\mathrm{1}−\boldsymbol{{cos}}\left(\boldsymbol{{x}}\right)\right)^{\mathrm{2}} }\boldsymbol{{dx}}=? \\ $$

Question Number 162109 Answers: 1 Comments: 0

Question Number 162103 Answers: 1 Comments: 0

Find: lim_(x→0) ((sinx - sin^(-1) x)/(sinhx - sinh^(-1) x)) = ?

$$\mathrm{Find}: \\ $$$$\underset{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\frac{\mathrm{sinx}\:-\:\mathrm{sin}^{-\mathrm{1}} \mathrm{x}}{\mathrm{sinhx}\:-\:\mathrm{sinh}^{-\mathrm{1}} \mathrm{x}}\:=\:? \\ $$

Question Number 162102 Answers: 0 Comments: 2

Solve for real numbers: 5^x + 4^(1/x) + 25^x ∙ 16^(1/x) = 2527

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{5}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{4}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \:+\:\mathrm{25}^{\boldsymbol{\mathrm{x}}} \:\centerdot\:\mathrm{16}^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}}} \:=\:\mathrm{2527} \\ $$

Question Number 162100 Answers: 1 Comments: 1

Question Number 162099 Answers: 0 Comments: 0

PROVE THAT Ω= ∫_0 ^( 1) (( Li_( 2) (x ). ln( x ))/x) dx =^? (( −π^( 4) )/(90)) −−−−−−−−−− Ω= ∫_0 ^( 1) ln (x )Σ_(n=1) ^∞ (( x^( n−1) )/n^( 2) ) dx = Σ_(n=1) ^∞ (1/n^( 2) ) ∫_0 ^( 1) x^( n−1) . ln(x ) dx = Σ_(n=1) ^∞ (1/n^( 2) ) {[ (x^( n) /n) ln( x )]_0 ^( 1) −(1/n) ∫_0 ^( 1) x^( n−1) dx} = Σ_(n=1) ^∞ ((−1)/n^( 4) ) = − ζ (4 ) = ((−π^( 4) )/( 90)) ■ m.n −−− M . N −−−

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathscr{PROVE}\:\:\:\mathscr{THAT}\:\: \\ $$$$\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{Li}_{\:\mathrm{2}} \:\left({x}\:\right).\:\mathrm{ln}\left(\:{x}\:\right)}{{x}}\:{dx}\:\overset{?} {=}\:\frac{\:−\pi^{\:\mathrm{4}} }{\mathrm{90}} \\ $$$$\:\:\:\:\:−−−−−−−−−− \\ $$$$\:\:\:\:\:\Omega=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \mathrm{ln}\:\left({x}\:\right)\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{x}^{\:{n}−\mathrm{1}} }{{n}^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \:{x}^{\:{n}−\mathrm{1}} .\:\mathrm{ln}\left({x}\:\right)\:{dx} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}^{\:\mathrm{2}} }\:\left\{\left[\:\frac{{x}^{\:{n}} }{{n}}\:\mathrm{ln}\left(\:{x}\:\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:{n}−\mathrm{1}} {dx}\right\} \\ $$$$\:\:\:\:\:\:\:\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{−\mathrm{1}}{{n}^{\:\mathrm{4}} }\:=\:−\:\zeta\:\left(\mathrm{4}\:\right)\:=\:\frac{−\pi^{\:\mathrm{4}} }{\:\mathrm{90}}\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:−−−\:\mathscr{M}\:.\:\mathscr{N}\:\:−−−\: \\ $$$$ \\ $$

Question Number 162139 Answers: 2 Comments: 0

Question Number 162138 Answers: 1 Comments: 0

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