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Question Number 141034    Answers: 0   Comments: 2

I don′t recover my old phone documents. please advise me in briefly how to restore my old phone documents in my new phone. plese help me.

$${I}\:{don}'{t}\:{recover}\:{my}\:{old}\:{phone}\:{documents}. \\ $$$${please}\:{advise}\:{me}\:{in}\:{briefly}\:{how}\:{to}\:{restore} \\ $$$$\:{my}\:{old}\:{phone}\:{documents}\:{in}\:{my}\:{new} \\ $$$$\:{phone}. \\ $$$$ \\ $$$${plese}\:{help}\:{me}. \\ $$

Question Number 141031    Answers: 0   Comments: 0

∫_(π/6) ^( π/3) (√(1+((cos^2 x)/(sin x)))) dx ?

$$\:\:\:\:\:\:\int_{\pi/\mathrm{6}} ^{\:\pi/\mathrm{3}} \:\sqrt{\mathrm{1}+\frac{\mathrm{cos}\:^{\mathrm{2}} {x}}{\mathrm{sin}\:{x}}}\:{dx}\:?\: \\ $$

Question Number 141030    Answers: 1   Comments: 1

→⟨Σ_(n=1) ^∞ (1/(9n^2 +3n)) =? ⟩←

$$\:\:\:\:\:\:\:\:\:\:\rightarrow\langle\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{1}}{\mathrm{9}{n}^{\mathrm{2}} +\mathrm{3}{n}}\:=?\:\rangle\leftarrow \\ $$

Question Number 141024    Answers: 3   Comments: 0

Given X=((√(√x))+(1/( (√(√x)))))^n . What is the coefficient of x^(5−(n/4) ?) propositions: a. 5((n!)/(10!)) b. ((n!)/(5!)) c. ((n),((10)) ) d. ((n),(5) )

$${Given}\:{X}=\left(\sqrt{\sqrt{{x}}}+\frac{\mathrm{1}}{\:\sqrt{\sqrt{{x}}}}\right)^{{n}} . \\ $$$${What}\:{is}\:{the}\:{coefficient}\:{of}\:{x}^{\mathrm{5}−\frac{\mathrm{n}}{\mathrm{4}}\:?} \\ $$$$\mathrm{propositions}: \\ $$$${a}.\:\:\:\:\mathrm{5}\frac{{n}!}{\mathrm{10}!} \\ $$$${b}.\:\:\:\:\frac{{n}!}{\mathrm{5}!} \\ $$$${c}.\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{10}}\end{pmatrix} \\ $$$${d}.\:\:\:\:\:\:\begin{pmatrix}{{n}}\\{\mathrm{5}}\end{pmatrix} \\ $$

Question Number 141017    Answers: 2   Comments: 1

Question Number 141004    Answers: 0   Comments: 0

Let 0 ≤ a,b < 1. Prove that (1/4)∙(((2−a)(2−b))/((1−a)(1−b))) ≥ ((4+a+b)/(4−a−b))

$$\mathrm{Let}\:\mathrm{0}\:\leqslant\:{a},{b}\:<\:\mathrm{1}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}−{a}\right)\left(\mathrm{2}−{b}\right)}{\left(\mathrm{1}−{a}\right)\left(\mathrm{1}−{b}\right)}\:\geqslant\:\frac{\mathrm{4}+{a}+{b}}{\mathrm{4}−{a}−{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140998    Answers: 1   Comments: 0

Question Number 140997    Answers: 0   Comments: 1

Let a,b,c ≥ 0. Prove that (1/8)∙(((2+a)(2+b)(2+c))/((1+a)(1+b)(1+c))) ≥ ((4−a−b−c)/(4+a+b+c))

$$\mathrm{Let}\:{a},{b},{c}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{8}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)\left(\mathrm{2}+{c}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)\left(\mathrm{1}+{c}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}−{c}}{\mathrm{4}+{a}+{b}+{c}}\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140996    Answers: 1   Comments: 0

Evaluation of :: Ω :=Σ_(n=0) ^∞ (((−1)^n )/(1+n^2 )) solution:: Ω:=1+Σ_(n=1) ^∞ (((−1)^n )/(n^2 −i^2 )) =(1/(2i)){Σ_(n=1) ^∞ (((−1)^n )/(n−i))−(((−1)^n )/(n+i))} :=1+(1/(2i)) (Φ−Ψ) where Φ:=Σ_(n=1) ^∞ (((−1)^n )/(n−i)) and Ψ :=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) Φ:=Σ_(n=1) ^∞ (1/(2n−k)) −Σ_(n=1) ^∞ (1/(2n−1−i)) :=(1/2){Σ_(n=1) ^∞ (1/(n−(i/2)))−Σ_(n=1) ^∞ (1/(n−((1+i)/2)))} :=(1/2){ψ(1−((1+i)/2))−ψ(1−(i/2))} :=(1/2)(ψ(((1−i)/2))−ψ(1−(i/2))).... Ψ:=Σ_(n=1) ^∞ (((−1)^n )/(n+i)) =Σ_(n=1) ^∞ (1/(2n+i))−Σ_(n=1) ^∞ (1/(2n−1+i)) :=(1/2){Σ_(n=1) ^∞ (1/(n+(i/2)))−Σ_(n=1) ^∞ (1/(n+((i−1)/2)))} :=(1/2)(ψ(((1+i)/2))−ψ(1+(i/2))) .... Φ−Ψ:=(1/2){ψ(((1−i)/2))−ψ(((1+i)/2))} +(1/2){ψ(1+(i/2))−ψ(1−(i/2))} :=(1/2)(−πcotπ(((1−i)/2)))+(1/2)((2/i)−πcot(π(i/2))) :=−(π/2)tan(((πi)/2))−(π/2)cot(((πi)/2))−i :=−i−(π/(sin(πi)))=−i−((2iπ)/(e^(−π) −e^π )) :=−i+πicsch(π) .... Ω :=1+(1/(2i))(−i+πicsch(π))=(1/2)+(π/2) csch(π) ... Ω:=(1/2)+(π/2) csch(π)....✓✓✓

$$ \\ $$$$\:\:\:\:\:\:\mathscr{E}{valuation}\:{of}\:::\:\Omega\::=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{1}+{n}^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:{solution}:: \\ $$$$\:\:\:\:\:\Omega:=\mathrm{1}+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}^{\mathrm{2}} −{i}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}{i}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}}−\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\right\} \\ $$$$\:\:\:\:\:\:\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\:\left(\Phi−\Psi\right)\:\:\:\:{where}\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}−{i}} \\ $$$$\:\:\:\:\:\:\:{and}\:\:\:\:\Psi\::=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\:\:\: \\ $$$$\:\:\:\:\:\:\:\Phi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−{k}}\:−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}−{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\frac{{i}}{\mathrm{2}}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}−\frac{\mathrm{1}+{i}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}−\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right).... \\ $$$$\:\:\:\:\Psi:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}+{i}}\:=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}+{i}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}+{i}} \\ $$$$\:\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left\{\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}}{\mathrm{2}}}−\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{{n}+\frac{{i}−\mathrm{1}}{\mathrm{2}}}\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)\right)\:.... \\ $$$$\:\:\:\:\:\Phi−\Psi:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)−\psi\left(\frac{\mathrm{1}+{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{2}}\left\{\psi\left(\mathrm{1}+\frac{{i}}{\mathrm{2}}\right)−\psi\left(\mathrm{1}−\frac{{i}}{\mathrm{2}}\right)\right\} \\ $$$$\:\:\:\:\:\:\::=\frac{\mathrm{1}}{\mathrm{2}}\left(−\pi{cot}\pi\left(\frac{\mathrm{1}−{i}}{\mathrm{2}}\right)\right)+\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{2}}{{i}}−\pi{cot}\left(\pi\frac{{i}}{\mathrm{2}}\right)\right) \\ $$$$\:\:\:\:\::=−\frac{\pi}{\mathrm{2}}{tan}\left(\frac{\pi{i}}{\mathrm{2}}\right)−\frac{\pi}{\mathrm{2}}{cot}\left(\frac{\pi{i}}{\mathrm{2}}\right)−{i} \\ $$$$\:\:\:\::=−{i}−\frac{\pi}{{sin}\left(\pi{i}\right)}=−{i}−\frac{\mathrm{2}{i}\pi}{{e}^{−\pi} −{e}^{\pi} } \\ $$$$\:\:\:\::=−{i}+\pi{icsch}\left(\pi\right)\:.... \\ $$$$\:\:\:\:\:\:\Omega\::=\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{i}}\left(−{i}+\pi{icsch}\left(\pi\right)\right)=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)\:... \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\Omega:=\frac{\mathrm{1}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}}\:{csch}\left(\pi\right)....\checkmark\checkmark\checkmark \\ $$$$\:\:\:\:\:\:\:\: \\ $$

Question Number 140988    Answers: 1   Comments: 0

∫((−csc^2 x)/((cscx+cotx)^3 ))dx

$$\int\frac{−{csc}^{\mathrm{2}} {x}}{\left({cscx}+{cotx}\right)^{\mathrm{3}} }{dx} \\ $$

Question Number 141012    Answers: 0   Comments: 0

s^2 (s+1)^2 +(c/2)s(s+1)^2 −(c^2 /2)(s+1) +(c^3 /2) = 0 solve for s in terms of 0<c<(2/(3(√3))) .

$$\:{s}^{\mathrm{2}} \left({s}+\mathrm{1}\right)^{\mathrm{2}} +\frac{{c}}{\mathrm{2}}{s}\left({s}+\mathrm{1}\right)^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\left({s}+\mathrm{1}\right) \\ $$$$\:\:+\frac{{c}^{\mathrm{3}} }{\mathrm{2}}\:=\:\mathrm{0}\:\:\:\:\:{solve}\:{for}\:{s}\:{in}\:{terms} \\ $$$${of}\:\:\:\:\mathrm{0}<{c}<\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$

Question Number 140982    Answers: 0   Comments: 3

convergence and value of Σ_(n=1) ^∞ (n^n /((n!)^2 ))

$${convergence}\:{and}\:{value}\:{of} \\ $$$$\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}^{{n}} }{\left({n}!\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$ \\ $$

Question Number 140981    Answers: 1   Comments: 0

Let α≠1 and α^(13) =1. If a=α+α^3 +α^4 +α^(−4) +α^(−3) + α^(−1) and b=α^2 +α^5 +α^6 +α^(−6) +α^(−5) +α^(−2) then the quadratic equation whose roots are a and b is (A) x^2 +x+3=0 (B) x^2 +x+4=0 (C) x^2 +x−3=0 (D) x^2 +x−4=0

$$\mathrm{Let}\:\alpha\neq\mathrm{1}\:\mathrm{and}\:\alpha^{\mathrm{13}} =\mathrm{1}.\:\mathrm{If}\:{a}=\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{−\mathrm{4}} +\alpha^{−\mathrm{3}} + \\ $$$$\alpha^{−\mathrm{1}} \:\mathrm{and}\:{b}=\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{−\mathrm{6}} +\alpha^{−\mathrm{5}} +\alpha^{−\mathrm{2}} \:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{quadratic}\:\mathrm{equation}\:\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:{a}\:\mathrm{and}\:{b}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{4}=\mathrm{0} \\ $$

Question Number 140978    Answers: 1   Comments: 0

calculate Σ_(n=0) ^∞ (1/((n!)^2 ))

$${calculate}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\left({n}!\right)^{\mathrm{2}} } \\ $$

Question Number 140977    Answers: 2   Comments: 0

find ∫_0 ^∞ (e^(−t(1+x^2 )) /(1+x^2 ))dx with t≥0

$${find}\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{t}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)} }{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:{with}\:{t}\geqslant\mathrm{0} \\ $$

Question Number 140976    Answers: 1   Comments: 0

find ∫_0 ^π (dx/((2−cosx−sinx)^2 ))

$${find}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\left(\mathrm{2}−{cosx}−{sinx}\right)^{\mathrm{2}} } \\ $$$$ \\ $$

Question Number 140974    Answers: 2   Comments: 3

If the equations x^2 −3x+a=0 and x^2 +ax−3=0 have a common root, then a possible value of a is (A) 3 (B) 1 (C) −2 (D) 2

$$\mathrm{If}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{x}^{\mathrm{2}} −\mathrm{3x}+{a}=\mathrm{0}\:\mathrm{and}\:\mathrm{x}^{\mathrm{2}} +{a}\mathrm{x}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root},\:\mathrm{then}\:\mathrm{a}\:\mathrm{possible}\:\mathrm{value}\:\mathrm{of}\:{a}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{C}\right)\:−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{2} \\ $$

Question Number 140973    Answers: 1   Comments: 0

If α and β are roots of the equation 2x^2 +ax+b=0, then one of the roots of the equation 2(αx+β)^2 + a(αx+β)+b=0 is (A) 0 (B) ((α+2b)/α^2 ) (C) ((aα+b)/(2α^2 )) (D) ((aα−2b)/(2α^2 ))

$$\mathrm{If}\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2x}^{\mathrm{2}} +{a}\mathrm{x}+{b}=\mathrm{0}, \\ $$$$\mathrm{then}\:\mathrm{one}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{2}\left(\alpha\mathrm{x}+\beta\right)^{\mathrm{2}} + \\ $$$${a}\left(\alpha\mathrm{x}+\beta\right)+{b}=\mathrm{0}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\frac{\alpha+\mathrm{2}{b}}{\alpha^{\mathrm{2}} } \\ $$$$\left(\mathrm{C}\right)\:\frac{{a}\alpha+{b}}{\mathrm{2}\alpha^{\mathrm{2}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\frac{{a}\alpha−\mathrm{2}{b}}{\mathrm{2}\alpha^{\mathrm{2}} } \\ $$

Question Number 140971    Answers: 0   Comments: 1

Question Number 140969    Answers: 0   Comments: 0

The roots of the equation x^2 −(m−3)x+m=0 are such that exactly one of them lies in the interval (1, 2). Then (A) 5<m<7 (B) m<10 (C) 2<m<5 (D) m>10

$$\mathrm{The}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{x}^{\mathrm{2}} −\left({m}−\mathrm{3}\right)\mathrm{x}+{m}=\mathrm{0}\:\mathrm{are} \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{exactly}\:\mathrm{one}\:\mathrm{of}\:\mathrm{them}\:\mathrm{lies}\:\mathrm{in}\:\mathrm{the}\:\mathrm{interval} \\ $$$$\left(\mathrm{1},\:\mathrm{2}\right).\:\mathrm{Then} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{5}<{m}<\mathrm{7}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:{m}<\mathrm{10} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{2}<{m}<\mathrm{5}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{m}>\mathrm{10} \\ $$

Question Number 140965    Answers: 0   Comments: 0

If a≠0 and a(l+m)^2 +2blm+c=0 and a(l+n)^2 + 2bln+c=0, then (A) mn=l^2 +c/a (B) lm=n^2 +c/a (C) ln=m^2 +c/a (D) mn=l^2 +bc/a

$$\mathrm{If}\:{a}\neq\mathrm{0}\:\mathrm{and}\:{a}\left({l}+{m}\right)^{\mathrm{2}} +\mathrm{2}{blm}+{c}=\mathrm{0}\:\mathrm{and}\:{a}\left({l}+{n}\right)^{\mathrm{2}} + \\ $$$$\mathrm{2}{bln}+{c}=\mathrm{0},\:\mathrm{then} \\ $$$$\left(\mathrm{A}\right)\:{mn}={l}^{\mathrm{2}} +{c}/{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:{lm}={n}^{\mathrm{2}} +{c}/{a} \\ $$$$\left(\mathrm{C}\right)\:{ln}={m}^{\mathrm{2}} +{c}/{a}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:{mn}={l}^{\mathrm{2}} +{bc}/{a} \\ $$

Question Number 140961    Answers: 2   Comments: 0

∫_0 ^( 1) ((ln (x+(√(1−x^2 ))))/x) dx =?

$$\:\underset{\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\:\frac{\mathrm{ln}\:\left({x}+\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }\right)}{{x}}\:{dx}\:=?\: \\ $$

Question Number 140959    Answers: 1   Comments: 0

Let a,b ≥ 0. Prove that (1/4)∙(((2+a)(2+b))/((1+a)(1+b))) ≥ ((4−a−b)/(4+a+b))

$$\mathrm{Let}\:{a},{b}\:\geqslant\:\mathrm{0}.\:\mathrm{Prove}\:\mathrm{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{4}}\centerdot\frac{\left(\mathrm{2}+{a}\right)\left(\mathrm{2}+{b}\right)}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+{b}\right)}\:\geqslant\:\frac{\mathrm{4}−{a}−{b}}{\mathrm{4}+{a}+{b}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 140958    Answers: 0   Comments: 0

find e^ (((−1 1)),((2 −1)) )

$$\mathrm{find}\:\mathrm{e}^{\begin{pmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{1}}\end{pmatrix}} \\ $$

Question Number 140956    Answers: 1   Comments: 0

.....advanced......calculus..... prove that: 𝛗:= ∫_(−∞) ^( ∞) ((sin^4 (x).cos^4 (x))/x^2 )dx=(π/(16)) m.n

$$\:\:\:\:\:\:\:\:\:\:.....{advanced}......{calculus}..... \\ $$$$\:\:\:\:\:{prove}\:{that}: \\ $$$$\:\:\boldsymbol{\phi}:=\:\int_{−\infty} ^{\:\infty} \frac{{sin}^{\mathrm{4}} \left({x}\right).{cos}^{\mathrm{4}} \left({x}\right)}{{x}^{\mathrm{2}} }{dx}=\frac{\pi}{\mathrm{16}} \\ $$$$\:\:{m}.{n} \\ $$

Question Number 140966    Answers: 1   Comments: 0

.......nice......calculus..... if Σ_(n=0) ^∞ (((√(cos (nπ))) )/((2n)!!)) = ω then Re(ω):=??

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:.......{nice}......{calculus}..... \\ $$$$\:\:\:\:\:{if}\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\sqrt{{cos}\:\left({n}\pi\right)}\:}{\left(\mathrm{2}{n}\right)!!}\:=\:\omega \\ $$$$\:\:\:\:\:\:\:{then}\:\:\:{Re}\left(\omega\right):=?? \\ $$

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