Question and Answers Forum

AllQuestion and Answers: Page 577

Question Number 160764 Answers: 0 Comments: 1

If log(((x^3 −y^3 )/(x^3 +y^3 ))), then (dy/dx)=?

$$\mathrm{If}\:\:\:\:\mathrm{log}\left(\frac{{x}^{\mathrm{3}} −{y}^{\mathrm{3}} }{{x}^{\mathrm{3}} +{y}^{\mathrm{3}} }\right),\:\mathrm{then}\:\frac{{dy}}{{dx}}=? \\ $$

Question Number 160762 Answers: 0 Comments: 5

sin 10+sin 20+sin 30+sin 40+∙∙∙∙+sin 360=?

$$\mathrm{sin}\:\mathrm{10}+\mathrm{sin}\:\mathrm{20}+\mathrm{sin}\:\mathrm{30}+\mathrm{sin}\:\mathrm{40}+\centerdot\centerdot\centerdot\centerdot+\mathrm{sin}\:\mathrm{360}=? \\ $$

Question Number 160746 Answers: 1 Comments: 0

Question Number 160744 Answers: 1 Comments: 1

(x^2 +x−12)^3 +(x^2 +3x−18)^2 = 9(x^2 −9)^2 x=?

$$\:\:\:\left(\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{12}\right)^{\mathrm{3}} +\left(\mathrm{x}^{\mathrm{2}} +\mathrm{3x}−\mathrm{18}\right)^{\mathrm{2}} =\:\mathrm{9}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{9}\right)^{\mathrm{2}} \\ $$$$\:\mathrm{x}=?\: \\ $$

Question Number 160739 Answers: 0 Comments: 3

nature de cette integrale quequ′en soit le reel α ∫_1 ^(+oo) t^α e^(−t) dt

$$\:{nature}\:{de}\:{cette}\:{integrale}\:{quequ}'{en}\:{soit}\:{le}\:{reel}\:\alpha \\ $$$$\int_{\mathrm{1}} ^{+{oo}} {t}^{\alpha} {e}^{−{t}} {dt} \\ $$

Question Number 160747 Answers: 1 Comments: 1

Question Number 160734 Answers: 0 Comments: 1

# Advanced Calculus # Φ = ∫_0 ^( 1) (((ln^ ( (1/(1− x)) ))/x) )^( 3) dx =^? 3 ( ζ (2 ) + ζ (3 )) −−−− solution−−−− Φ =^(I.B.P) [ (( 1)/(2x^( 2) )) ln^( 3) ( 1−x)]_0 ^1 +(3/2) ∫_0 ^( 1) (( ln^( 2) (1− x ))/(x^( 2) (1 − x ))) dx = (1/2) lim_( ξ →1^(− ) ) ((ln^( 3) ( 1− ξ ))/ξ^( 2) ) +(3/2)[∫_0 ^( 1) (( ln^( 2) ( 1− x ))/x)dx = 2 ζ (3)] + (3/2)[∫_0 ^( 1) (( ln^( 2) ( 1−x))/x^( 2) ) dx = (π^( 2) /3) = 2ζ (2 )] +(3/2)∫_0 ^( 1) (( ln^( 2) (1− x))/(1−x)) dx} =(1/2) lim_( ξ →1^( −) ) {((ln^( 3) ( 1−ξ ))/ξ^( 2) ) −ln^( 3) (1− ξ ) } +(3/2) (2ζ (3 )) +(3/2) ( 2ζ (2 )) = 3( ζ (3 ) + 3ζ (2 ) ) ■ m.n

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:#\:\mathrm{Advanced}\:\:\:\mathrm{Calculus}\:# \\ $$$$\:\:\:\:\:\:\:\:\Phi\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\frac{{ln}^{\:} \:\left(\:\frac{\mathrm{1}}{\mathrm{1}−\:{x}}\:\:\right)}{{x}}\:\right)^{\:\mathrm{3}} {dx}\:\overset{?} {=}\:\mathrm{3}\:\left(\:\zeta\:\left(\mathrm{2}\:\right)\:+\:\zeta\:\left(\mathrm{3}\:\right)\right) \\ $$$$\:\:\:\:\:\:−−−−\:\:{solution}−−−− \\ $$$$\:\:\:\:\:\:\:\:\Phi\:\overset{\mathrm{I}.\mathrm{B}.\mathrm{P}} {=}\:\left[\:\frac{\:\mathrm{1}}{\mathrm{2}{x}^{\:\mathrm{2}} }\:{ln}^{\:\mathrm{3}} \left(\:\mathrm{1}−{x}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\frac{\mathrm{3}}{\mathrm{2}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \:\left(\mathrm{1}−\:{x}\:\right)}{{x}^{\:\mathrm{2}} \:\left(\mathrm{1}\:−\:{x}\:\right)}\:{dx} \\ $$$$\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:{lim}_{\:\xi\:\rightarrow\mathrm{1}^{−\:} } \:\frac{{ln}^{\:\mathrm{3}} \left(\:\mathrm{1}−\:\xi\:\right)}{\xi^{\:\mathrm{2}} }\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{3}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\:\mathrm{1}−\:{x}\:\right)}{{x}}{dx}\:=\:\mathrm{2}\:\zeta\:\left(\mathrm{3}\right)\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\:\frac{\mathrm{3}}{\mathrm{2}}\left[\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\:\mathrm{1}−{x}\right)}{{x}^{\:\mathrm{2}} }\:{dx}\:=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{3}}\:=\:\mathrm{2}\zeta\:\left(\mathrm{2}\:\right)\right]\: \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−\:{x}\right)}{\mathrm{1}−{x}}\:{dx}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\:{lim}_{\:\xi\:\rightarrow\mathrm{1}^{\:−} } \left\{\frac{{ln}^{\:\mathrm{3}} \left(\:\mathrm{1}−\xi\:\right)}{\xi^{\:\mathrm{2}} }\:\:−{ln}^{\:\mathrm{3}} \left(\mathrm{1}−\:\xi\:\right)\:\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\frac{\mathrm{3}}{\mathrm{2}}\:\left(\mathrm{2}\zeta\:\left(\mathrm{3}\:\right)\right)\:\:+\frac{\mathrm{3}}{\mathrm{2}}\:\left(\:\mathrm{2}\zeta\:\left(\mathrm{2}\:\right)\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\:\mathrm{3}\left(\:\:\:\zeta\:\left(\mathrm{3}\:\right)\:+\:\mathrm{3}\zeta\:\left(\mathrm{2}\:\right)\:\:\right)\:\:\:\:\:\:\:\blacksquare\:\:\:{m}.{n}\:\:\: \\ $$$$ \\ $$

Question Number 160733 Answers: 0 Comments: 0

Question Number 160755 Answers: 0 Comments: 0

Question Number 160753 Answers: 1 Comments: 1

Question Number 160752 Answers: 1 Comments: 0

solve (d^2 y/dx^2 )−y=x^2 sin3x

$${solve} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−{y}={x}^{\mathrm{2}} {sin}\mathrm{3}{x} \\ $$

Question Number 160719 Answers: 1 Comments: 2

Question Number 160807 Answers: 0 Comments: 0

−1≤a_0 ≤b_0 ≤c_0 ≤1 ∀n∈N a_(n+1) =∫_(−1) ^1 min(x,b_n ,c_n )dx b_(n+1) =∫_(−1) ^1 mil(x,a_n ,c_n )dx c_(n+1) =∫_(−1) ^1 max(x,b_n ,a_n )dx mil(a,b,c) est le terme median de (a,b,c) nature de (a_n ),(b_n ),(c_n )

$$−\mathrm{1}\leqslant{a}_{\mathrm{0}} \leqslant{b}_{\mathrm{0}} \leqslant{c}_{\mathrm{0}} \leqslant\mathrm{1} \\ $$$$\forall{n}\in\mathbb{N}\: \\ $$$${a}_{{n}+\mathrm{1}} =\int_{−\mathrm{1}} ^{\mathrm{1}} {min}\left({x},{b}_{{n}} ,{c}_{{n}} \right){dx} \\ $$$${b}_{{n}+\mathrm{1}} =\int_{−\mathrm{1}} ^{\mathrm{1}} {mil}\left({x},{a}_{{n}} ,{c}_{{n}} \right){dx} \\ $$$${c}_{{n}+\mathrm{1}} =\int_{−\mathrm{1}} ^{\mathrm{1}} {max}\left({x},{b}_{{n}} ,{a}_{{n}} \right){dx} \\ $$$${mil}\left({a},{b},{c}\right)\:{est}\:{le}\:{terme}\:{median}\:{de}\:\left({a},{b},{c}\right) \\ $$$${nature}\:{de}\:\left({a}_{{n}} \right),\left({b}_{{n}} \right),\left({c}_{{n}} \right) \\ $$$$ \\ $$

Question Number 160806 Answers: 1 Comments: 0

Question Number 160712 Answers: 0 Comments: 0

Solve for real numbers: (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) y-1 ; (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) z-1 (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) x-1 n ∈ (0 ; ∞) fixed

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3n}^{\mathrm{2}} }{\mathrm{4n}^{\mathrm{2}} }\right)^{\mathrm{4}} =\:\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{y}-\mathrm{1}\:\:;\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3n}^{\mathrm{2}} }{\mathrm{4n}^{\mathrm{2}} }\right)^{\mathrm{4}} =\:\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{z}-\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{3n}^{\mathrm{2}} }{\mathrm{4n}^{\mathrm{2}} }\right)^{\mathrm{4}} =\:\frac{\mathrm{2}}{\mathrm{n}}\:\mathrm{x}-\mathrm{1} \\ $$$$\mathrm{n}\:\in\:\left(\mathrm{0}\:;\:\infty\right)\:\:\boldsymbol{\mathrm{fixed}} \\ $$

Question Number 160711 Answers: 2 Comments: 0

Be p a prime number , arbitrary. Solve on positive integers (x;y;z) { ((xy + z^2 = 3p + 4)),((x + yz^2 = p + 4)) :}

$$\mathrm{Be}\:\:\boldsymbol{\mathrm{p}}\:\:\mathrm{a}\:\mathrm{prime}\:\mathrm{number}\:,\:\mathrm{arbitrary}. \\ $$$$\mathrm{Solve}\:\mathrm{on}\:\mathrm{positive}\:\mathrm{integers}\:\:\left(\boldsymbol{\mathrm{x}};\boldsymbol{\mathrm{y}};\boldsymbol{\mathrm{z}}\right) \\ $$$$\begin{cases}{\mathrm{xy}\:+\:\mathrm{z}^{\mathrm{2}} \:=\:\mathrm{3p}\:+\:\mathrm{4}}\\{\mathrm{x}\:+\:\mathrm{yz}^{\mathrm{2}} \:=\:\mathrm{p}\:+\:\mathrm{4}}\end{cases} \\ $$

Question Number 160707 Answers: 0 Comments: 0