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For what value of k the roots of the equation ((x^2 −2x)/(4x−1))=((k−1)/(k+1)) will have same value but with opposite symbol(like x=a and −a) i mean the two valuea of x will be this type x=2 and −2(both 2 but opposite symbols)

$${For}\:{what}\:{value}\:{of}\:\:{k}\:\:{the}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\frac{{x}^{\mathrm{2}} −\mathrm{2}{x}}{\mathrm{4}{x}−\mathrm{1}}=\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}} \\ $$$${will}\:{have}\:{same}\:{value}\:{but}\:\:{with}\:{opposite}\:{symbol}\left({like}\:{x}={a}\:{and}\:−{a}\right) \\ $$$${i}\:{mean}\:{the}\:{two}\:{valuea}\:{of}\:{x}\:{will}\:{be}\:{this}\:{type} \\ $$$${x}=\mathrm{2}\:{and}\:−\mathrm{2}\left({both}\:\mathrm{2}\:{but}\:{opposite}\:{symbols}\right) \\ $$

Question Number 222295 Answers: 1 Comments: 0

Question Number 222292 Answers: 1 Comments: 0

∫_(−∞) ^∞ sech(z) sech(z−a) dz

$$ \\ $$$$\:\:\:\int_{−\infty} ^{\infty} \mathrm{sech}\left({z}\right)\:\mathrm{sech}\left({z}−{a}\right)\:{dz} \\ $$$$ \\ $$

Question Number 222288 Answers: 1 Comments: 0

Prove that: 1 + 2 + 3 + ... + n = ((n∙(n + 1))/2)

$$\mathrm{Prove}\:\mathrm{that}: \\ $$$$\mathrm{1}\:+\:\mathrm{2}\:+\:\mathrm{3}\:+\:...\:+\:\boldsymbol{\mathrm{n}}\:=\:\frac{\mathrm{n}\centerdot\left(\mathrm{n}\:+\:\mathrm{1}\right)}{\mathrm{2}} \\ $$

Question Number 222296 Answers: 0 Comments: 0

(1+x^4 )y′−x^3 y = x^5 −x^3 +2x+1

$$\left(\mathrm{1}+{x}^{\mathrm{4}} \right){y}'−{x}^{\mathrm{3}} {y}\:=\:{x}^{\mathrm{5}} −{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1} \\ $$

Question Number 222284 Answers: 1 Comments: 0

y=(8^x /((in8)^3 )) find (d^6 y/dx^6 )

$$\boldsymbol{\mathrm{y}}=\frac{\mathrm{8}^{\boldsymbol{\mathrm{x}}} }{\left(\boldsymbol{\mathrm{in}}\mathrm{8}\right)^{\mathrm{3}} } \\ $$$$\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{6}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{6}} } \\ $$

Question Number 222280 Answers: 0 Comments: 1

y=3x^(2024) −18x^(2020) +5x^(47) −8 find (d^(2025) y/dx^(2025) )

$$\boldsymbol{\mathrm{y}}=\mathrm{3}\boldsymbol{\mathrm{x}}^{\mathrm{2024}} −\mathrm{18}\boldsymbol{\mathrm{x}}^{\mathrm{2020}} +\mathrm{5}\boldsymbol{\mathrm{x}}^{\mathrm{47}} −\mathrm{8} \\ $$$$\boldsymbol{\mathrm{find}}\:\frac{\boldsymbol{\mathrm{d}}^{\mathrm{2025}} \boldsymbol{\mathrm{y}}}{\boldsymbol{\mathrm{dx}}^{\mathrm{2025}} } \\ $$

Question Number 222279 Answers: 2 Comments: 0

(a^2 −b^2 )sin θ+2abcos θ=a^2 +b^2 tan θ=??

$$\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)\mathrm{sin}\:\theta+\mathrm{2}{ab}\mathrm{cos}\:\theta={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\mathrm{tan}\:\theta=?? \\ $$

Question Number 222276 Answers: 0 Comments: 0

Prove; Σ_(k = 1) ^∞ (k/(sinh(πk))) = ((Γ((1/4))^4 − 8π^2 )/(32π^3 ))

$$ \\ $$$$\:\mathrm{Prove};\:\underset{{k}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{k}}{\mathrm{sinh}\left(\pi{k}\right)}\:=\:\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{4}} −\:\mathrm{8}\pi^{\mathrm{2}} }{\mathrm{32}\pi^{\mathrm{3}} }\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222275 Answers: 1 Comments: 0

Prove ; ∫_(−π) ^( π) ((z sin(z) )/((1 + z + (√(1 + z^2 )))(√(3 + sin^2 (z))))) dz = ζ(2)

$$ \\ $$$$\:\:\mathrm{Prove}\:;\:\int_{−\pi} ^{\:\pi} \:\frac{{z}\:\mathrm{sin}\left({z}\right)\:}{\left(\mathrm{1}\:+\:{z}\:+\:\sqrt{\mathrm{1}\:+\:{z}^{\mathrm{2}} }\right)\sqrt{\mathrm{3}\:+\:\mathrm{sin}^{\mathrm{2}} \left({z}\right)}}\:{dz}\:=\:\zeta\left(\mathrm{2}\right)\:\:\:\:\:\:\:\: \\ $$$$ \\ $$

Question Number 222271 Answers: 2 Comments: 0

y=(((1+(√5))/2))^(10) ,Prove:y=((123+55(√5))/2)

$${y}=\left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)^{\mathrm{10}} ,\mathrm{Prove}:{y}=\frac{\mathrm{123}+\mathrm{55}\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Question Number 222261 Answers: 1 Comments: 0

lim_(x→0) ((tan(x^2 +4x))/(sin(9x^2 +x))) No L′ho^ pital′s rule allowed!

$$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{tan}\left({x}^{\mathrm{2}} +\mathrm{4}{x}\right)}{\mathrm{sin}\left(\mathrm{9}{x}^{\mathrm{2}} +{x}\right)} \\ $$$$\mathrm{No}\:\mathrm{L}'\mathrm{h}\hat {\mathrm{o}pital}'\mathrm{s}\:\mathrm{rule}\:\mathrm{allowed}! \\ $$

Question Number 222249 Answers: 3 Comments: 0

Prove:∀n∈Z^+ ,1^3 +2^3 +…+n^3 =(1+2+…+n)^2

$$\mathrm{Prove}:\forall{n}\in\mathbb{Z}^{+} ,\mathrm{1}^{\mathrm{3}} +\mathrm{2}^{\mathrm{3}} +\ldots+{n}^{\mathrm{3}} =\left(\mathrm{1}+\mathrm{2}+\ldots+{n}\right)^{\mathrm{2}} \\ $$

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