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Question Number 156423    Answers: 1   Comments: 2

Question Number 156422    Answers: 1   Comments: 2

Question Number 156421    Answers: 0   Comments: 2

Question Number 156419    Answers: 1   Comments: 1

x is positive integer number can you check if Q=((((x+2)^4 −x^4 ))^(1/3) is a natural number

$$\mathrm{x}\:\mathrm{is}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{number}\: \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{check}\:\mathrm{if}\:\mathrm{Q}=\sqrt[{\mathrm{3}}]{\left(\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{4}} −\mathrm{x}^{\mathrm{4}} \right.} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{natural}\:\mathrm{number} \\ $$

Question Number 156413    Answers: 2   Comments: 0

Question Number 156404    Answers: 0   Comments: 3

x^4 +bx^2 +cx+d=0 let cx=m+px^2 +x^4 ⇒ 2x^4 +(b+p)x^2 +m+d=0 x^2 =−(((b+p)/4))±(√((((b+p)/4))^2 −(((m+d)/2)))) (p−b)x^2 −2cx+m−d=0 x=(c/(p−b))±(√(((c/(p−b)))^2 −(((m−d)/(p−b))))) ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −8(p−b)^2 (m+d))/(16))) +((2c^2 )/(b−p))∓(√((4c^4 −4c^2 (p−b)(m−d))/((p−b)^2 ))) +m−d=0 Just if m=d ⇒ (((b^2 −p^2 ))/4)±(√(((b^2 −p^2 )^2 −16d(p−b))/(16))) +((4c^2 )/(b−p))=0 ⇒ (((b^2 −p^2 )/4)+((4c^2 )/(b−p)))^2 +d(p−b)=(((b^2 −p^2 )^2 )/(16)) ⇒ ((16c^4 )/((b−p)^2 ))+(d+2c^2 )(b+p)=0 let b−p=z z^2 (2b−z)=−((16c^4 )/(2c^2 +d)) z^3 −2bz^2 −k=0 (k>0) if b=−1 , d=0, c→−c z^3 +2z^2 −8c^2 =0 let z=((2c)/t) 8c^3 +8c^2 t−8c^2 t^3 =0 ⇒ t^3 −t=c back to square one.

$$\:\:\mathrm{x}^{\mathrm{4}} +\mathrm{bx}^{\mathrm{2}} +\mathrm{cx}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{cx}=\mathrm{m}+\mathrm{px}^{\mathrm{2}} +\mathrm{x}^{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{2x}^{\mathrm{4}} +\left(\mathrm{b}+\mathrm{p}\right)\mathrm{x}^{\mathrm{2}} +\mathrm{m}+\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} =−\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)\pm\sqrt{\left(\frac{\mathrm{b}+\mathrm{p}}{\mathrm{4}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{m}+\mathrm{d}}{\mathrm{2}}\right)} \\ $$$$\left(\mathrm{p}−\mathrm{b}\right)\mathrm{x}^{\mathrm{2}} −\mathrm{2cx}+\mathrm{m}−\mathrm{d}=\mathrm{0} \\ $$$$\mathrm{x}=\frac{\mathrm{c}}{\mathrm{p}−\mathrm{b}}\pm\sqrt{\left(\frac{\mathrm{c}}{\mathrm{p}−\mathrm{b}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{m}−\mathrm{d}}{\mathrm{p}−\mathrm{b}}\right)} \\ $$$$\Rightarrow \\ $$$$\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)}{\mathrm{4}}\pm\sqrt{\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{8}\left(\mathrm{p}−\mathrm{b}\right)^{\mathrm{2}} \left(\mathrm{m}+\mathrm{d}\right)}{\mathrm{16}}} \\ $$$$+\frac{\mathrm{2c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}\mp\sqrt{\frac{\mathrm{4c}^{\mathrm{4}} −\mathrm{4c}^{\mathrm{2}} \left(\mathrm{p}−\mathrm{b}\right)\left(\mathrm{m}−\mathrm{d}\right)}{\left(\mathrm{p}−\mathrm{b}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:+\mathrm{m}−\mathrm{d}=\mathrm{0} \\ $$$${Just}\:{if}\:\:{m}={d}\:\:\Rightarrow \\ $$$$\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)}{\mathrm{4}}\pm\sqrt{\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{16d}\left(\mathrm{p}−\mathrm{b}\right)}{\mathrm{16}}} \\ $$$$+\frac{\mathrm{4c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}=\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\left(\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{4c}^{\mathrm{2}} }{\mathrm{b}−\mathrm{p}}\right)^{\mathrm{2}} +\mathrm{d}\left(\mathrm{p}−\mathrm{b}\right)=\frac{\left(\mathrm{b}^{\mathrm{2}} −\mathrm{p}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{16}} \\ $$$$\Rightarrow\:\frac{\mathrm{16c}^{\mathrm{4}} }{\left(\mathrm{b}−\mathrm{p}\right)^{\mathrm{2}} }+\left(\mathrm{d}+\mathrm{2c}^{\mathrm{2}} \right)\left(\mathrm{b}+\mathrm{p}\right)=\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{b}−\mathrm{p}=\mathrm{z} \\ $$$$\mathrm{z}^{\mathrm{2}} \left(\mathrm{2b}−\mathrm{z}\right)=−\frac{\mathrm{16c}^{\mathrm{4}} }{\mathrm{2c}^{\mathrm{2}} +\mathrm{d}} \\ $$$$\mathrm{z}^{\mathrm{3}} −\mathrm{2bz}^{\mathrm{2}} −\mathrm{k}=\mathrm{0}\:\:\:\:\left(\mathrm{k}>\mathrm{0}\right) \\ $$$$\mathrm{if}\:\mathrm{b}=−\mathrm{1}\:,\:\mathrm{d}=\mathrm{0},\:\mathrm{c}\rightarrow−\mathrm{c} \\ $$$$\mathrm{z}^{\mathrm{3}} +\mathrm{2z}^{\mathrm{2}} −\mathrm{8c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{z}=\frac{\mathrm{2c}}{\mathrm{t}} \\ $$$$\mathrm{8c}^{\mathrm{3}} +\mathrm{8c}^{\mathrm{2}} \mathrm{t}−\mathrm{8c}^{\mathrm{2}} \mathrm{t}^{\mathrm{3}} =\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{t}^{\mathrm{3}} −\mathrm{t}=\mathrm{c} \\ $$$$\mathrm{back}\:\mathrm{to}\:\mathrm{square}\:\mathrm{one}. \\ $$$$ \\ $$$$ \\ $$

Question Number 156403    Answers: 0   Comments: 1

find two numbers between (1/3)and (3/4)

$$\mathrm{find}\:\mathrm{two}\:\mathrm{numbers}\:\mathrm{between}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{3}}\mathrm{and}\:\frac{\mathrm{3}}{\mathrm{4}} \\ $$

Question Number 156400    Answers: 0   Comments: 1

Question Number 156384    Answers: 0   Comments: 0

Question Number 156381    Answers: 0   Comments: 1

∫_0 ^( (𝛑/2)) (x^2 /(sin(x))) dx

$$\int_{\mathrm{0}} ^{\:\frac{\boldsymbol{\pi}}{\mathrm{2}}} \:\frac{\boldsymbol{{x}}^{\mathrm{2}} }{\boldsymbol{{sin}}\left(\boldsymbol{{x}}\right)}\:\boldsymbol{{dx}} \\ $$

Question Number 156386    Answers: 1   Comments: 0

φ (n )= Σ_(k=1) ^n (−1 )^( k−1) ((( n)),(( k)) ) H_( k) Find the value of : Σ_(n=1) ^∞ (−1)^( n−1) φ ( n^( 2) ) =?

$$ \\ $$$$\:\:\:\:\:\:\:\:\phi\:\left({n}\:\right)=\:\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(−\mathrm{1}\:\right)^{\:{k}−\mathrm{1}} \begin{pmatrix}{\:{n}}\\{\:{k}}\end{pmatrix}\:\mathrm{H}_{\:{k}} \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{Find}\:\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\right)^{\:{n}−\mathrm{1}} \:\phi\:\left(\:{n}^{\:\mathrm{2}} \right)\:=? \\ $$

Question Number 156373    Answers: 1   Comments: 0

A. lim_(x→+∞) Σ_(k=1) ^n [tan((kπ)/(2n))]

$$ \\ $$$$\:\:\mathrm{A}.\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{1}} {\overset{\mathrm{n}} {\sum}}\left[\mathrm{tan}\frac{{k}\pi}{\mathrm{2n}}\right] \\ $$

Question Number 156369    Answers: 1   Comments: 2

solve by betta function ∫_0 ^(2π) sin^5 (z) dz

$${solve}\:{by}\:{betta}\:{function}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} {sin}^{\mathrm{5}} \left({z}\right)\:{dz} \\ $$

Question Number 156367    Answers: 0   Comments: 0

lim→oo nΣ_(k=0) ^(2n−1) (1/k^2 )

$$\:{lim}\rightarrow{oo}\:{n}\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{{k}^{\mathrm{2}} } \\ $$

Question Number 156363    Answers: 0   Comments: 0

∫_0 ^(π/2) t^(2n) sin^(2n) tdt

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {t}^{\mathrm{2}{n}} {sin}^{\mathrm{2}{n}} {tdt} \\ $$

Question Number 156362    Answers: 1   Comments: 0

Question Number 156361    Answers: 1   Comments: 2

Solve in R (1/( (√(x^2 - 1)))) =1− (2/x)

$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{1}}}\:=\mathrm{1}−\:\frac{\mathrm{2}}{\mathrm{x}} \\ $$

Question Number 156359    Answers: 0   Comments: 0

if a;b;c>0 and ab+bc+ca=3 then prove that: (((a^3 + 1)(b^3 + 1)(c^3 + 1)))^(1/3) ≥ 2

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}>\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{ab}+\mathrm{bc}+\mathrm{ca}=\mathrm{3} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\sqrt[{\mathrm{3}}]{\left(\mathrm{a}^{\mathrm{3}} \:+\:\mathrm{1}\right)\left(\mathrm{b}^{\mathrm{3}} \:+\:\mathrm{1}\right)\left(\mathrm{c}^{\mathrm{3}} \:+\:\mathrm{1}\right)}\:\geqslant\:\mathrm{2} \\ $$

Question Number 156339    Answers: 1   Comments: 0

∫_0 ^1 (x^(2n) /(arcsinx))dx=...???

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}{n}} }{{arcsinx}}{dx}=...??? \\ $$

Question Number 156338    Answers: 1   Comments: 2

Solve in R (1/( (√(x^2 - 1)))) = (2/x) - 1

$$\mathrm{Solve}\:\mathrm{in}\:\mathbb{R} \\ $$$$\frac{\mathrm{1}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{1}}}\:=\:\frac{\mathrm{2}}{\mathrm{x}}\:-\:\mathrm{1} \\ $$$$ \\ $$

Question Number 156335    Answers: 1   Comments: 1

Question Number 156333    Answers: 1   Comments: 0

Question Number 156395    Answers: 0   Comments: 0

A car is moving along a straight road level when the driver see a boy crossing the road 1.5m when the driver immediately applies the brakes which produces constant retardation in the first second. After applying the brakes the car travels 25m and in the next second it traveles 15m. i) find the retardation in m/s^2 ii) show that the car comes to rest at the point where it started

$${A}\:{car}\:{is}\:{moving}\:{along}\:{a}\:{straight}\:{road}\:{level}\:{when}\: \\ $$$${the}\:{driver}\:{see}\:{a}\:{boy}\:{crossing}\:{the}\:{road}\:\mathrm{1}.\mathrm{5}{m}\:{when}\:{the}\:{driver} \\ $$$${immediately}\:{applies}\:{the}\:{brakes}\:{which}\:{produces} \\ $$$$\:{constant}\:{retardation}\:{in}\:{the}\:{first}\:{second}.\:{After}\:{applying}\:{the}\:{brakes} \\ $$$${the}\:{car}\:{travels}\:\mathrm{25}{m}\:{and}\:{in}\:{the}\:{next}\:{second}\:{it} \\ $$$${traveles}\:\mathrm{15}{m}.\: \\ $$$$\left.{i}\right)\:{find}\:{the}\:{retardation}\:{in}\:{m}/{s}^{\mathrm{2}} \\ $$$$\left.{ii}\right)\:{show}\:{that}\:{the}\:{car}\:{comes}\:{to}\:{rest}\:{at}\: \\ $$$${the}\:{point}\:{where}\:{it}\:{started} \\ $$

Question Number 156330    Answers: 0   Comments: 0

Question Number 156328    Answers: 1   Comments: 4

Question Number 156321    Answers: 0   Comments: 1

cos^4 ((π/9))+cos^4 (((2π)/9))+cos^4 (((3π)/9))+cos^4 (((4π)/9))=?

$$\:\:\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\pi}{\mathrm{9}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{2}\pi}{\mathrm{9}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{3}\pi}{\mathrm{9}}\right)+\mathrm{cos}\:^{\mathrm{4}} \left(\frac{\mathrm{4}\pi}{\mathrm{9}}\right)=? \\ $$

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