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Question Number 202589 Answers: 1 Comments: 2

Ω = Σ_(k=1) ^∞ ((1/k) + 2ln(1−(1/(2k))))=?

$$ \\ $$$$\:\:\Omega\:=\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{k}}\:+\:\mathrm{2}{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}{k}}\right)\right)=?\:\:\:\:\: \\ $$$$ \\ $$$$ \\ $$

Question Number 202564 Answers: 0 Comments: 0

Find the transformation of Fourier: z(t)=cos(2πF_1 t)+cos(2πF_2 t)

$${Find}\:{the}\:{transformation}\:{of}\:{Fourier}: \\ $$$${z}\left({t}\right)={cos}\left(\mathrm{2}\pi{F}_{\mathrm{1}} {t}\right)+{cos}\left(\mathrm{2}\pi{F}_{\mathrm{2}} {t}\right) \\ $$

Question Number 202551 Answers: 0 Comments: 1

Σ_(i=1) ^n (x+y)_i =(x+y)_1 +(x+y)_2 +...(x+y)_n =x_1 +y_1 +x_2 +y_2 +...+x_n +y_n please it′s correct ?

$$\underset{\mathrm{i}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{i}} =\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{1}} +\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{2}} +...\left(\mathrm{x}+\mathrm{y}\right)_{\mathrm{n}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{x}_{\mathrm{1}} +\mathrm{y}_{\mathrm{1}} +\mathrm{x}_{\mathrm{2}} +\mathrm{y}_{\mathrm{2}} +...+\mathrm{x}_{\mathrm{n}} +\mathrm{y}_{\mathrm{n}} \: \\ $$$$\mathrm{please}\:\mathrm{it}'\mathrm{s}\:\mathrm{correct}\:? \\ $$$$ \\ $$

Question Number 202543 Answers: 1 Comments: 1

Question Number 202540 Answers: 1 Comments: 0

If (b/(a + b)) = ((3a − b − c)/(2b + c − a)) = ((3c − a)/(2a − b + 3c)) [a + b + c ≠ 0] then show that a = b = c.

$$\mathrm{If}\:\frac{{b}}{{a}\:+\:{b}}\:=\:\frac{\mathrm{3}{a}\:−\:{b}\:−\:{c}}{\mathrm{2}{b}\:+\:{c}\:−\:{a}}\:=\:\frac{\mathrm{3}{c}\:−\:{a}}{\mathrm{2}{a}\:−\:{b}\:+\:\mathrm{3}{c}}\: \\ $$$$\left[{a}\:+\:{b}\:+\:{c}\:\neq\:\mathrm{0}\right]\:\mathrm{then}\:\mathrm{show}\:\mathrm{that}\:{a}\:=\:{b}\:=\:{c}. \\ $$

Question Number 202536 Answers: 2 Comments: 0

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Question Number 202535 Answers: 1 Comments: 0

Solve for x ((x^3 +3x−3))^(1/3) +((−x^3 −3x+5))^(1/3) =2 (An alteration of Q#202500)

$$\boldsymbol{\mathrm{Solve}}\:\boldsymbol{\mathrm{for}}\:\boldsymbol{{x}} \\ $$$$\sqrt[{\mathrm{3}}]{{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{3}}\:+\sqrt[{\mathrm{3}}]{−{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{5}}\:=\mathrm{2} \\ $$$$\left(\mathrm{An}\:\mathrm{alteration}\:\mathrm{of}\:\mathrm{Q}#\mathrm{202500}\right) \\ $$

Question Number 202532 Answers: 0 Comments: 1

please help

$${please}\:{help} \\ $$

Question Number 202530 Answers: 1 Comments: 0

Question Number 202522 Answers: 0 Comments: 0

Question Number 202514 Answers: 0 Comments: 0

Question Number 202511 Answers: 0 Comments: 0

Question Number 202500 Answers: 4 Comments: 0

Question Number 202497 Answers: 3 Comments: 0

If the difference of the roots of x^2 + 2px + q = 0 is equal to the difference of the roots of x^2 + 2qx + p = 0 [p ≠ q] then show that p + q + 1 = 0.

$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of} \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{px}\:+\:{q}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{the}\: \\ $$$$\mathrm{difference}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:+\:\mathrm{2}{qx}\:+\:{p}\:=\:\mathrm{0}\:\left[{p}\:\neq\:{q}\right]\:\mathrm{then}\:\mathrm{show} \\ $$$$\mathrm{that}\:{p}\:+\:{q}\:+\:\mathrm{1}\:=\:\mathrm{0}. \\ $$

Question Number 202490 Answers: 2 Comments: 2

Question Number 202485 Answers: 0 Comments: 3

Question Number 202477 Answers: 3 Comments: 0

If x = (((√(a^2 + b^2 )) + (√(a^2 − b^2 )))/( (√(a^2 + b^2 )) − (√(a^2 − b^2 )))) then show that b^2 x^2 − 2a^2 x + b^2 = 0.

$$\mathrm{If}\:{x}\:=\:\frac{\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:+\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}{\:\sqrt{{a}^{\mathrm{2}} \:+\:{b}^{\mathrm{2}} }\:−\:\sqrt{{a}^{\mathrm{2}} \:−\:{b}^{\mathrm{2}} }}\:\mathrm{then}\:\mathrm{show}\:\mathrm{that} \\ $$$${b}^{\mathrm{2}} {x}^{\mathrm{2}} \:−\:\mathrm{2}{a}^{\mathrm{2}} {x}\:+\:{b}^{\mathrm{2}} \:=\:\mathrm{0}. \\ $$

Question Number 202473 Answers: 0 Comments: 0

Find: 1. Σ_(n=1) ^∞ ((2^n ∙ n!)/((2n)!)) 2. Σ_(n=1) ^∞ (((x + 5)^n )/(3^(n+1) ∙ n ∙ ln(n)))

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\mathrm{2}^{\boldsymbol{\mathrm{n}}} \:\centerdot\:\mathrm{n}!}{\left(\mathrm{2n}\right)!}\:\:\:\:\:\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{\left(\mathrm{x}\:+\:\mathrm{5}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{3}^{\boldsymbol{\mathrm{n}}+\mathrm{1}} \:\centerdot\:\mathrm{n}\:\centerdot\:\mathrm{ln}\left(\mathrm{n}\right)} \\ $$

Question Number 202468 Answers: 1 Comments: 0

Find: 1. Σ_(n=1) ^( ∞) ((16)/(16n^2 − 8n − 3)) = ? 2. Σ_(n=1) ^( ∞) (((−1)^n )/(2n^3 )) = ?

$$\mathrm{Find}: \\ $$$$\mathrm{1}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\mathrm{16}}{\mathrm{16n}^{\mathrm{2}} \:−\:\mathrm{8n}\:−\:\mathrm{3}}\:=\:?\:\:\:\:\:\mathrm{2}.\:\underset{\boldsymbol{\mathrm{n}}=\mathrm{1}} {\overset{\:\infty} {\sum}}\:\frac{\left(−\mathrm{1}\right)^{\boldsymbol{\mathrm{n}}} }{\mathrm{2n}^{\mathrm{3}} }\:=\:? \\ $$

Question Number 202466 Answers: 1 Comments: 0

lim_(x→+∞) ((x(√(ln (x^2 +1))))/(1+e^(x−3) ))

$$\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\frac{{x}\sqrt{\mathrm{ln}\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{1}+{e}^{{x}−\mathrm{3}} } \\ $$

Question Number 202464 Answers: 0 Comments: 0

Question Number 202462 Answers: 1 Comments: 0

Question Number 202459 Answers: 3 Comments: 0

If the difference of two roots of x^2 − lx + m = 0 is 1 then prove that l^2 + 4m^2 = (1 + 2m)^2 .

$$\mathrm{If}\:\mathrm{the}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$${x}^{\mathrm{2}} \:−\:{lx}\:+\:{m}\:=\:\mathrm{0}\:\mathrm{is}\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${l}^{\mathrm{2}} \:+\:\mathrm{4}{m}^{\mathrm{2}} \:=\:\left(\mathrm{1}\:+\:\mathrm{2}{m}\right)^{\mathrm{2}} \:. \\ $$

Question Number 202449 Answers: 0 Comments: 0

Question Number 202448 Answers: 2 Comments: 0

Question Number 202447 Answers: 2 Comments: 0

rationnalise le denominateur de x = (((2)^(1/(3 )) −1)/(1−^3 (√2)+^3 (√4)))

$$\mathrm{rationnalise}\:\mathrm{le}\:\mathrm{denominateur}\:\mathrm{de}\: \\ $$$$\mathrm{x}\:=\:\frac{\sqrt[{\mathrm{3}\:}]{\mathrm{2}}−\mathrm{1}}{\mathrm{1}−^{\mathrm{3}} \sqrt{\mathrm{2}}+^{\mathrm{3}} \sqrt{\mathrm{4}}} \\ $$

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