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Question Number 166391 Answers: 2 Comments: 0

solve in R (√(x.⌊x⌋)) − (√(⌊x⌋)) = 1 −−−−−−−

$$ \\ $$$$\:\:{solve}\:{in}\:\:\mathbb{R} \\ $$$$\:\:\:\sqrt{{x}.\lfloor{x}\rfloor}\:−\:\sqrt{\lfloor{x}\rfloor}\:=\:\mathrm{1} \\ $$$$\:\:−−−−−−− \\ $$

Question Number 166379 Answers: 1 Comments: 0

soit{_(u_1 =4) ^(u_o =1) ∀nεN montrer par reccurence que: U_(n+2) =2U_(n+1) −U_n

$${soit}\left\{_{{u}_{\mathrm{1}} =\mathrm{4}} ^{{u}_{{o}} =\mathrm{1}} \:\:\forall{n}\epsilon{N}\right. \\ $$$${montrer}\:{par}\:{reccurence}\:\:{que}: \\ $$$${U}_{{n}+\mathrm{2}} =\mathrm{2}{U}_{{n}+\mathrm{1}} −{U}_{{n}} \\ $$

Question Number 166377 Answers: 1 Comments: 0

Given A= ((((1/2) (1/2))),(( a b)) ) . If A^3 = A^2 then 2a−3b=?

$$\:\:\mathrm{Given}\:\mathrm{A}=\begin{pmatrix}{\frac{\mathrm{1}}{\mathrm{2}}\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}}}\\{\:\:\mathrm{a}\:\:\:\:\:\:\:\:\mathrm{b}}\end{pmatrix}\:.\:\mathrm{If}\:\mathrm{A}^{\mathrm{3}} =\:\mathrm{A}^{\mathrm{2}} \\ $$$$\:\mathrm{then}\:\mathrm{2a}−\mathrm{3b}=? \\ $$

Question Number 166376 Answers: 0 Comments: 0

Question Number 166378 Answers: 1 Comments: 0

8x^3 −6x+1=0 solve for x

$$\:\mathrm{8}{x}^{\mathrm{3}} −\mathrm{6}{x}+\mathrm{1}=\mathrm{0} \\ $$$$\:{solve}\:{for}\:{x} \\ $$

Question Number 166373 Answers: 0 Comments: 0

prove 𝛗=∫_0 ^( 1) (( ln^( 2) (1−x^( 2) ) )/x^( 2) ) dx =(π^( 2) /3) −4ln^( 2) (2) −−− solution (technical method) −−− 𝛗= ∫_0 ^( 1) ln^( 2) (1−x^( 2) )d(1−(1/x)) = [(1−(1/x))ln^( 2) (1−x^( 2) )]_0 ^1 +4∫_0 ^( 1) (1−(1/x))((xln(1−x^( 2) ))/(1−x^( 2) ))dx = −4∫_0 ^( 1) ((ln(1−x^( 2) ))/(1+x)) dx = −4∫_0 ^( 1) ((ln(1+x))/(1+x))dx −4∫_0 ^( 1) ((ln(1−x)dx)/(1+x)) = −2ln^( 2) (2) −4 ( −(π^( 2) /(12)) +(1/2)ln^( 2) (2)) ∴ 𝛗= (π^( 2) /3) −4ln^( 2) (2) ■ m.n

$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{prove}\:\: \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:{ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)\:}{{x}^{\:\mathrm{2}} }\:{dx}\:=\frac{\pi^{\:\mathrm{2}} }{\mathrm{3}}\:−\mathrm{4}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:−−−\:\:{solution}\:\left({technical}\:{method}\right)\:−−− \\ $$$$\:\:\:\:\boldsymbol{\phi}=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{\:\mathrm{2}} \right){d}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right) \\ $$$$\:\:\:\:\:\:\:\:=\:\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right){ln}^{\:\mathrm{2}} \left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)\right]_{\mathrm{0}} ^{\mathrm{1}} +\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \left(\mathrm{1}−\frac{\mathrm{1}}{{x}}\right)\frac{{xln}\left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)}{\mathrm{1}−{x}^{\:\mathrm{2}} }{dx} \\ $$$$\:\:\:\:\:\:\:=\:−\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}^{\:\mathrm{2}} \right)}{\mathrm{1}+{x}}\:{dx}\: \\ $$$$\:\:\:\:\:\:\:=\:−\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}+{x}\right)}{\mathrm{1}+{x}}{dx}\:−\mathrm{4}\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{{ln}\left(\mathrm{1}−{x}\right){dx}}{\mathrm{1}+{x}} \\ $$$$\:\:\:\:\:\:=\:−\mathrm{2}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right)\:−\mathrm{4}\:\left(\:−\frac{\pi^{\:\mathrm{2}} }{\mathrm{12}}\:+\frac{\mathrm{1}}{\mathrm{2}}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right)\right) \\ $$$$\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\boldsymbol{\phi}=\:\frac{\pi^{\:\mathrm{2}} }{\mathrm{3}}\:−\mathrm{4}{ln}^{\:\mathrm{2}} \left(\mathrm{2}\right)\:\:\:\:\:\:\:\:\:\:\blacksquare\:\:{m}.{n}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Question Number 166371 Answers: 0 Comments: 2