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Question Number 176637 Answers: 0 Comments: 0

Question Number 176636 Answers: 2 Comments: 2

{ ((p^3 +q^3 =r^2 )),((p^3 +r^3 =q^2 )),((q^3 +r^3 =p^2 )) :} ⇒20pqr =?

$$\:\begin{cases}{{p}^{\mathrm{3}} +{q}^{\mathrm{3}} ={r}^{\mathrm{2}} }\\{{p}^{\mathrm{3}} +{r}^{\mathrm{3}} ={q}^{\mathrm{2}} }\\{{q}^{\mathrm{3}} +{r}^{\mathrm{3}} ={p}^{\mathrm{2}} }\end{cases} \\ $$$$\:\Rightarrow\mathrm{20}{pqr}\:=? \\ $$

Question Number 176610 Answers: 1 Comments: 0

Question Number 176607 Answers: 1 Comments: 0

Question Number 176603 Answers: 2 Comments: 6

look the anser

$${look}\:{the}\:{anser} \\ $$

Question Number 176595 Answers: 2 Comments: 0

Question Number 176594 Answers: 0 Comments: 1

𝛗=∫_0 ^( 1) (( ( tanh^( −1) (x))^2 )/((1+x )^( 2) )) dx = ? ≺ solution ≻ note : tanh^( −1) (x)=− (1/2) ln(((1−x)/(1+x))) 𝛗= (1/4)∫_0 ^( 1) (( ln^( 2) (((1−x)/(1+x)) ))/((1+x )^( 2) )) dx =^(((1−x)/(1+x)) = t) (1/8)∫_0 ^( 1) ln^( 2) (t )dt =(1/8_ ) { [t.ln^( 2) (t)]_0 ^( 1) −2∫_0 ^( 1) ln(t)dt} =− (1/4) ∫_0 ^( 1) ln(t)dt= (1/4) ◂ m.n ▶

$$ \\ $$$$\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\left(\:{tanh}^{\:−\mathrm{1}} \left({x}\right)\right)^{\mathrm{2}} }{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx}\:=\:?\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\prec\:\:\:{solution}\:\:\succ \\ $$$$\:\:\:\:\:{note}\::\:\:{tanh}^{\:−\mathrm{1}} \left({x}\right)=−\:\frac{\mathrm{1}}{\mathrm{2}}\:{ln}\left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\right) \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\:\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:\right)}{\left(\mathrm{1}+{x}\:\right)^{\:\mathrm{2}} }\:{dx} \\ $$$$\:\:\:\:\:\:\:\overset{\frac{\mathrm{1}−{x}}{\mathrm{1}+{x}}\:=\:{t}} {=}\:\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}^{\:\mathrm{2}} \left({t}\:\right){dt} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}_{\:} }\:\left\{\:\left[{t}.{ln}^{\:\mathrm{2}} \left({t}\right)\right]_{\mathrm{0}} ^{\:\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=−\:\frac{\mathrm{1}}{\mathrm{4}}\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}\left({t}\right){dt}=\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\:\:\:\:\:\blacktriangleleft\:{m}.{n}\:\blacktriangleright\: \\ $$

Question Number 176592 Answers: 2 Comments: 0

Solve the equation: 2^x + 3^x − 4^x + 6^x − 9^x = 1

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\mathrm{2}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{3}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{4}^{\boldsymbol{\mathrm{x}}} \:+\:\mathrm{6}^{\boldsymbol{\mathrm{x}}} \:−\:\mathrm{9}^{\boldsymbol{\mathrm{x}}} \:=\:\mathrm{1} \\ $$

Question Number 176589 Answers: 1 Comments: 0

Question Number 176598 Answers: 1 Comments: 0

x^3 +(1/x^3 )=1 (((x^5 +(1/x^5 ))^3 −1)/(x^5 +(1/x^5 )))=? Q#176387 reposted for a new answer.

$${x}^{\mathrm{3}} +\frac{\mathrm{1}}{{x}^{\mathrm{3}} }=\mathrm{1} \\ $$$$\frac{\left({x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }\right)^{\mathrm{3}} −\mathrm{1}}{{x}^{\mathrm{5}} +\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}=? \\ $$$${Q}#\mathrm{176387}\:{reposted}\:{for}\:{a}\:{new}\:{answer}. \\ $$

Question Number 176581 Answers: 1 Comments: 0

Given { ((sin a+sin b=((√2)/2))),((cos a+cos b=((√6)/2))) :} for a,b real numbers. Evaluate sin (a+b). (A)((√3)/2) (D) −((√3)/2) (B) (2/( (√3))) (E)−(2/( (√3))) (C) ((√3)/4)

$$\:\:\mathrm{Given}\:\begin{cases}{\mathrm{sin}\:\mathrm{a}+\mathrm{sin}\:\mathrm{b}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{cos}\:\mathrm{a}+\mathrm{cos}\:\mathrm{b}=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}}}\end{cases} \\ $$$$\:\mathrm{for}\:\mathrm{a},\mathrm{b}\:\mathrm{real}\:\mathrm{numbers}.\:\mathrm{Evaluate} \\ $$$$\:\mathrm{sin}\:\left(\mathrm{a}+\mathrm{b}\right). \\ $$$$\:\left(\mathrm{A}\right)\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\:\left(\mathrm{B}\right)\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\left(\mathrm{E}\right)−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\left(\mathrm{C}\right)\:\frac{\sqrt{\mathrm{3}}}{\mathrm{4}} \\ $$

Question Number 176580 Answers: 1 Comments: 0

4^(x^2 −2x+2) −2^(x^2 −2x+3) +2=2^(x^2 −2x+2) x=?

$$\mathrm{4}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}} −\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{3}} +\mathrm{2}=\mathrm{2}^{{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}} \\ $$$${x}=? \\ $$

Question Number 176571 Answers: 1 Comments: 0

lim_(x→1) ((lnx)/(1+lnx−1))=?

$$\underset{{x}\rightarrow\mathrm{1}} {\mathrm{lim}}\frac{{lnx}}{\mathrm{1}+{lnx}−\mathrm{1}}=? \\ $$

Question Number 176570 Answers: 2 Comments: 0

(1) ∫^(π/2) _(π/3) ((1+sinx)/(cosx)) dx=?

$$\left(\mathrm{1}\right)\:\:\underset{\frac{\pi}{\mathrm{3}}} {\int}^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{sinx}}{{cosx}}\:{dx}=? \\ $$

Question Number 176566 Answers: 0 Comments: 0

−−−− calculate: Φ = Σ_(n=0) ^( ∞) (( 1)/((2n+1 ).e^( 4n+2) )) = ? where ” e ” is euler number. ≺ solution ≻ Φ = Σ_(n=0) ^∞ (1/e^( 4n+2) ) ∫_0 ^( 1) x^( 2n) dx = (1/e^( 2) ) ∫_0 ^( 1) Σ_(n=0) ^∞ ((( x^2 )/e^( 4) ) )^( n) dx = (1/e^( 2) ) ∫_0 ^( 1) (( 1)/(1− ((x/e^( 2) ) )^( 2) )) dx=(1/(2e^( 2) )) ∫_0 ^( 1) (1/(1−(x/e^( 2) ))) +(1/(1+(x/e^( 2) )))dx = (1/2) ln ( ((1+(1/e^( 2) ))/(1−(1/e^( 2) ))) ) = tanh^( −1) ((( 1)/e^( 2) ) ) ∴ Φ = coth^( −1) ( e^( 2) ) ■ m.n

$$−−−− \\ $$$$\:\:{calculate}:\:\:\:\:\Phi\:=\:\underset{{n}=\mathrm{0}} {\overset{\:\infty} {\sum}}\:\frac{\:\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\:\right).{e}^{\:\mathrm{4}{n}+\mathrm{2}} }\:=\:? \\ $$$$\:\:\:\:\:\:\:\:\:\:{where}\:\:''\:\:{e}\:\:''\:\:{is}\:\:{euler}\:{number}. \\ $$$$\:\:\:\:\:\:\prec\:\:\:{solution}\:\:\succ \\ $$$$\:\:\:\:\:\:\Phi\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{{e}^{\:\mathrm{4}{n}+\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{n}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\:\:{x}^{\mathrm{2}} }{{e}^{\:\mathrm{4}} }\:\right)^{\:{n}} {dx}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{1}}{\mathrm{1}−\:\left(\frac{{x}}{{e}^{\:\mathrm{2}} }\:\right)^{\:\mathrm{2}} }\:{dx}=\frac{\mathrm{1}}{\mathrm{2}{e}^{\:\mathrm{2}} }\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\mathrm{1}}{\mathrm{1}−\frac{{x}}{{e}^{\:\mathrm{2}} }}\:+\frac{\mathrm{1}}{\mathrm{1}+\frac{{x}}{{e}^{\:\mathrm{2}} }}{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\:\:\frac{\mathrm{1}}{\mathrm{2}}\:\:{ln}\:\left(\:\:\frac{\mathrm{1}+\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }}{\mathrm{1}−\frac{\mathrm{1}}{{e}^{\:\mathrm{2}} }}\:\:\right)\:\:\:\:=\:{tanh}^{\:−\mathrm{1}} \left(\frac{\:\mathrm{1}}{{e}^{\:\mathrm{2}} }\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\Phi\:=\:{coth}^{\:−\mathrm{1}} \left(\:{e}^{\:\mathrm{2}} \right)\:\:\:\:\:\:\blacksquare\:{m}.{n}\:\:\:\: \\ $$$$\:\:\:\: \\ $$

Question Number 176565 Answers: 0 Comments: 0

Question Number 176564 Answers: 0 Comments: 0

Question Number 176563 Answers: 1 Comments: 0

is there an iOS version for this app? please

$$\mathrm{is}\:\mathrm{there}\:\mathrm{an}\:\mathrm{iOS}\:\mathrm{version}\:\mathrm{for}\:\mathrm{this}\:\mathrm{app}?\:\mathrm{please} \\ $$

Question Number 176555 Answers: 1 Comments: 2

ABCD−convex quadrilateral M∈Int(ABCD) , F−area , s−semiperimetr a , b , c−sides. Prove that: ((MA^4 )/b) + ((MB^4 )/c^4 ) + ((MC^4 )/d) + ((MD^4 )/a) ≥ ((2F^2 )/s)

$$\mathrm{ABCD}−\mathrm{convex}\:\mathrm{quadrilateral} \\ $$$$\mathrm{M}\in\mathrm{Int}\left(\mathrm{ABCD}\right)\:,\:\mathrm{F}−\mathrm{area}\:,\:\mathrm{s}−\mathrm{semiperimetr} \\ $$$$\mathrm{a}\:,\:\mathrm{b}\:,\:\mathrm{c}−\mathrm{sides}.\:\mathrm{Prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{MA}^{\mathrm{4}} }{\mathrm{b}}\:+\:\frac{\mathrm{MB}^{\mathrm{4}} }{\mathrm{c}^{\mathrm{4}} }\:+\:\frac{\mathrm{MC}^{\mathrm{4}} }{\mathrm{d}}\:+\:\frac{\mathrm{MD}^{\mathrm{4}} }{\mathrm{a}}\:\geqslant\:\frac{\mathrm{2F}^{\mathrm{2}} }{\mathrm{s}} \\ $$

Question Number 176554 Answers: 2 Comments: 0

x + (1/x) =ϕ (Golden ratio) then x^( 2000) +(( 1)/x^( 2000) ) = ?

$$ \\ $$$$\:\:\:\:{x}\:+\:\frac{\mathrm{1}}{{x}}\:\:=\varphi\:\:\left({Golden}\:{ratio}\right) \\ $$$$\:\:\:{then} \\ $$$$\:\:\:\:\:\:\:\:\:{x}^{\:\mathrm{2000}} \:+\frac{\:\mathrm{1}}{{x}^{\:\mathrm{2000}} }\:=\:? \\ $$$$ \\ $$

Question Number 176576 Answers: 0 Comments: 0