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Question Number 157314    Answers: 0   Comments: 1

lim_(x→∞) ((3 ln (1+5tan (4/x)))/(x (1−cos (6/x)))) =?

$$\:\:\:\:\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\frac{\mathrm{3}\:\mathrm{ln}\:\left(\mathrm{1}+\mathrm{5tan}\:\frac{\mathrm{4}}{{x}}\right)}{{x}\:\left(\mathrm{1}−\mathrm{cos}\:\frac{\mathrm{6}}{{x}}\right)}\:=? \\ $$

Question Number 157309    Answers: 3   Comments: 0

∫ (dx/( (√x)+x(√(x+1)))) =?

$$\:\:\int\:\frac{{dx}}{\:\sqrt{{x}}+{x}\sqrt{{x}+\mathrm{1}}}\:=? \\ $$

Question Number 157292    Answers: 1   Comments: 6

Question Number 157283    Answers: 1   Comments: 0

Question Number 157278    Answers: 3   Comments: 0

Question Number 157272    Answers: 2   Comments: 0

for solving equation which one we use ⇒ and =, i mean where we use ⇒ and where we use, = and where we use one of them to consider wrong.

$${for}\:{solving}\:{equation}\:{which}\:{one}\:{we}\:{use} \\ $$$$\Rightarrow\:{and}\:=,\:{i}\:{mean}\:{where}\:{we}\:{use}\:\Rightarrow\:{and}\:{where}\:{we}\:{use},\:= \\ $$$${and}\:{where}\:{we}\:{use}\:{one}\:{of}\:{them}\:{to} \\ $$$${consider}\:{wrong}. \\ $$

Question Number 157268    Answers: 1   Comments: 0

prove that ∫(x^2 /((xsin x+cos x)^2 ))dx=−((xsec x)/(xsin x+cos x))+tan x+c

$${prove}\:{that} \\ $$$$\int\frac{{x}^{\mathrm{2}} }{\left({x}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}\right)^{\mathrm{2}} }{dx}=−\frac{{x}\mathrm{sec}\:{x}}{{x}\mathrm{sin}\:{x}+\mathrm{cos}\:{x}}+\mathrm{tan}\:{x}+{c} \\ $$

Question Number 157267    Answers: 0   Comments: 3

(1+x^2 )y′′−2xy′+2y=x

$$\:\left(\mathrm{1}+{x}^{\mathrm{2}} \right){y}''−\mathrm{2}{xy}'+\mathrm{2}{y}={x}\: \\ $$

Question Number 157265    Answers: 1   Comments: 0

Solve for real numbers: sin(x) + cos(x) + sec(x)∙csc(x)=2+(√2)

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\mathrm{sin}\left(\mathrm{x}\right)\:+\:\mathrm{cos}\left(\mathrm{x}\right)\:+\:\mathrm{sec}\left(\mathrm{x}\right)\centerdot\mathrm{csc}\left(\mathrm{x}\right)=\mathrm{2}+\sqrt{\mathrm{2}} \\ $$

Question Number 157258    Answers: 0   Comments: 2

Solve for real numbers: (1/(sin^(2k) (x))) + (1/(cos^(2k) (x))) = 8 ; k∈Z

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}\boldsymbol{\mathrm{k}}} \left(\mathrm{x}\right)}\:+\:\frac{\mathrm{1}}{\mathrm{cos}^{\mathrm{2}\boldsymbol{\mathrm{k}}} \left(\mathrm{x}\right)}\:=\:\mathrm{8}\:\:\:;\:\:\:\mathrm{k}\in\mathbb{Z} \\ $$

Question Number 157257    Answers: 1   Comments: 0

∫ ((sin^6 x+cos^5 x)/(sin^2 x cos^2 x)) dx

$$\int\:\frac{\mathrm{sin}\:^{\mathrm{6}} {x}+\mathrm{cos}\:^{\mathrm{5}} {x}}{\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx} \\ $$

Question Number 157251    Answers: 1   Comments: 0

# Nice Mathematics # ...calculation ... Ω :=∫_0 ^( 1) (( tanh^( −1) ((√( x)) ))/x) dx =^? (( π^( 2) )/4) −−−−−−−−−−−−− Ω :=^((√x) = t) 2∫_0 ^( 1) (( tanh^( −1) (t ))/t) dt :=^({tanh^( −1) (t )= (1/2) ln( ((1+t)/(1−t)) ) }) ∫_0 ^( 1) ((ln( 1+t )− ln(1−t ))/t) dt : = −Li_( 2) (−1 ) + Li_( 2) (1 ) :=^( {Li_( 2) (z )= Σ_(n=1) ^∞ (( z^( n) )/n^( 2) ) }) η (2) + ζ (2) := (π^( 2) /(12)) + (π^( 2) /6) = (( π^( 2) )/( 4)) ■ m.n

$$ \\ $$$$\:\:\:\:\:#\:\mathrm{Nice}\:\mathrm{Mathematics}\:# \\ $$$$\:\:\:\:\:\:\:...{calculation}\:... \\ $$$$\:\:\:\:\:\:\:\:\Omega\::=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{tanh}^{\:−\mathrm{1}} \:\left(\sqrt{\:{x}}\:\right)}{{x}}\:{dx}\:\overset{?} {=}\:\frac{\:\pi^{\:\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:−−−−−−−−−−−−− \\ $$$$\:\:\:\:\Omega\::\overset{\sqrt{{x}}\:=\:{t}} {=}\:\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{tanh}^{\:−\mathrm{1}} \:\left({t}\:\right)}{{t}}\:{dt} \\ $$$$\:\:\:\:\:\:\:\:\::\overset{\left\{{tanh}^{\:−\mathrm{1}} \:\left({t}\:\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\:\frac{\mathrm{1}+{t}}{\mathrm{1}−{t}}\:\right)\:\right\}} {=}\:\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\:\mathrm{1}+{t}\:\right)−\:{ln}\left(\mathrm{1}−{t}\:\right)}{{t}}\:{dt} \\ $$$$\:\:\:\::\:\:=\:\:−\mathrm{Li}_{\:\mathrm{2}} \:\left(−\mathrm{1}\:\right)\:+\:\mathrm{Li}_{\:\mathrm{2}} \:\left(\mathrm{1}\:\right) \\ $$$$\:\:\:\:\::\overset{\:\left\{\mathrm{Li}_{\:\mathrm{2}} \:\left({z}\:\right)=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\:{z}^{\:{n}} }{{n}^{\:\mathrm{2}} }\:\right\}} {=}\:\:\eta\:\left(\mathrm{2}\right)\:+\:\zeta\:\left(\mathrm{2}\right)\: \\ $$$$\:\:\:\:\::=\:\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{12}}\:+\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{6}}\:\:=\:\frac{\:\pi^{\:\mathrm{2}} }{\:\mathrm{4}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 157247    Answers: 3   Comments: 0

F(x,y)=x^2 −2xy+6y^2 −12x+2y+45 find x &y such that F(x,y) minimum

$${F}\left({x},{y}\right)={x}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{6}{y}^{\mathrm{2}} −\mathrm{12}{x}+\mathrm{2}{y}+\mathrm{45} \\ $$$${find}\:{x}\:\&{y}\:{such}\:{that}\:{F}\left({x},{y}\right)\:{minimum} \\ $$

Question Number 157241    Answers: 0   Comments: 0

Question Number 157231    Answers: 2   Comments: 0

SOLVE : ⌊ x ⌋ + ⌊2x ⌋ +⌊ 3x ⌋= 1 −−−−−−−−−−

$$ \\ $$$$\:\:\:\:\:\:\mathcal{SOLVE}\:: \\ $$$$ \\ $$$$\:\:\:\:\:\lfloor\:{x}\:\rfloor\:+\:\lfloor\mathrm{2}{x}\:\rfloor\:+\lfloor\:\mathrm{3}{x}\:\rfloor=\:\mathrm{1} \\ $$$$−−−−−−−−−− \\ $$$$ \\ $$

Question Number 157433    Answers: 0   Comments: 0

Question Number 157230    Answers: 1   Comments: 0

(3x+1)^(100) Find this max Koeffitcient

$$\left(\mathrm{3x}+\mathrm{1}\right)^{\mathrm{100}} \:\: \\ $$$$\mathrm{Find}\:\mathrm{this}\:\mathrm{max}\:\mathrm{Koeffitcient} \\ $$

Question Number 157438    Answers: 1   Comments: 0

find all subgroups of : a) grup (Z_6 , +) b) grup (Z_6 −{0}, ×)

$$\mathrm{find}\:\mathrm{all}\:\mathrm{subgroups}\:\mathrm{of}\:: \\ $$$$\left.\mathrm{a}\right)\:\mathrm{grup}\:\left(\mathrm{Z}_{\mathrm{6}} \:,\:+\right) \\ $$$$\left.\mathrm{b}\right)\:\mathrm{grup}\:\left(\mathrm{Z}_{\mathrm{6}} \:−\left\{\mathrm{0}\right\},\:×\right) \\ $$

Question Number 157436    Answers: 1   Comments: 0

if x;y;z≥0 then: 2^(x+y+z) + 2 ≥ 2^x + 2^y + 2^z

$$\mathrm{if}\:\:\mathrm{x};\mathrm{y};\mathrm{z}\geqslant\mathrm{0}\:\:\mathrm{then}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{y}}+\boldsymbol{\mathrm{z}}} \:+\:\mathrm{2}\:\geqslant\:\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{y}}} \:+\:\mathrm{2}^{\boldsymbol{\mathrm{z}}} \\ $$$$ \\ $$

Question Number 157435    Answers: 1   Comments: 0

Question Number 157225    Answers: 0   Comments: 2

Question Number 157219    Answers: 0   Comments: 0

Σ_(0<n) (((−1)^(n−1) n)/(sinh(πn)))=(1/(4π)) prove

$$\underset{\mathrm{0}<\boldsymbol{\mathrm{n}}} {\sum}\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}−\mathrm{1}} \boldsymbol{\mathrm{n}}}{\boldsymbol{\mathrm{sinh}}\left(\pi\boldsymbol{\mathrm{n}}\right)}=\frac{\mathrm{1}}{\mathrm{4}\pi}\:\:\:\:{prove} \\ $$

Question Number 157254    Answers: 1   Comments: 0

Show that ∫_0 ^1 (1/((x+1)(x+2)))dx = ln((4/3))

$$\mathrm{Show}\:\mathrm{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)}{dx}\:=\:\mathrm{ln}\left(\frac{\mathrm{4}}{\mathrm{3}}\right) \\ $$

Question Number 157199    Answers: 1   Comments: 0

What is the general expression of the divergence(divV^⇢ )

$${What}\:{is}\:{the}\:{general}\:{expression}\:{of}\:{the}\:{divergence}\left({div}\overset{\dashrightarrow} {{V}}\right) \\ $$

Question Number 157227    Answers: 1   Comments: 0

x(3sin((√x))−2(√x))=sin^3 ((√x))

$$\boldsymbol{\mathrm{x}}\left(\mathrm{3}\boldsymbol{\mathrm{sin}}\left(\sqrt{\boldsymbol{\mathrm{x}}}\right)−\mathrm{2}\sqrt{\boldsymbol{\mathrm{x}}}\right)=\boldsymbol{\mathrm{sin}}^{\mathrm{3}} \left(\sqrt{\boldsymbol{\mathrm{x}}}\right) \\ $$

Question Number 157191    Answers: 1   Comments: 0

lim_(x→0) ((sin x − x + (1/6) x^3 )/x^5 ) = ? ( Without L′Hospital , Taylor or Maclaurin Series ) .

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\:\frac{\mathrm{sin}\:{x}\:−\:{x}\:+\:\frac{\mathrm{1}}{\mathrm{6}}\:{x}^{\mathrm{3}} }{{x}^{\mathrm{5}} }\:\:\:=\:\:? \\ $$$$\left(\:{Without}\:\:{L}'{Hospital}\:,\:{Taylor}\:\:{or}\:\:{Maclaurin}\:\:{Series}\:\right)\:. \\ $$

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