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Question Number 196115    Answers: 1   Comments: 0

Question Number 196111    Answers: 1   Comments: 0

Question Number 196107    Answers: 1   Comments: 0

help, please ! lim_(x → (𝛑/4)) ((cos((𝛑/4) −x)−tan x)/(1−sin((𝛑/4)+x))) = ???

$$ \\ $$$$\mathrm{help},\:\mathrm{please}\:! \\ $$$$\underset{{x}\:\rightarrow\:\frac{\boldsymbol{\pi}}{\mathrm{4}}} {{lim}}\:\frac{{cos}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}\:−{x}\right)−{tan}\:{x}}{\mathrm{1}−{sin}\left(\frac{\boldsymbol{\pi}}{\mathrm{4}}+{x}\right)}\:=\:???\: \\ $$

Question Number 196103    Answers: 0   Comments: 0

(d^n /dx^n )[𝚪(x)] Hence what is (d^(1/2) /dx^(1/2) )[𝚪(x)]

$$\frac{\boldsymbol{\mathrm{d}}^{\boldsymbol{\mathrm{n}}} }{\boldsymbol{\mathrm{dx}}^{\boldsymbol{\mathrm{n}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right]\:\:\boldsymbol{\mathrm{Hence}}\:\boldsymbol{\mathrm{what}}\:\boldsymbol{\mathrm{is}}\:\frac{\boldsymbol{\mathrm{d}}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\boldsymbol{\mathrm{dx}}^{\frac{\mathrm{1}}{\mathrm{2}}} }\left[\boldsymbol{\Gamma}\left(\boldsymbol{\mathrm{x}}\right)\right] \\ $$

Question Number 196102    Answers: 0   Comments: 0

Question Number 196099    Answers: 0   Comments: 0

Question Number 196098    Answers: 1   Comments: 0

Question Number 196097    Answers: 0   Comments: 0

Question Number 196094    Answers: 1   Comments: 0

Question Number 196134    Answers: 3   Comments: 0

Question Number 196086    Answers: 0   Comments: 0

Answer to the question “196008” Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1) Ans) according “de moivre” sin(2n+1)α= (((2n+1)),(( 1)) )(cosα)^(2n) sinα− (((2n+1)),(( 3)) )(cosα)^(2n−2) (sinα)^3 +.... =(cosα)^(2n) (sinα)[ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α+...] for “α_k =((kπ)/(2n+1)) ; 1≤k≤n ⇒sin(2n+1)α_k =0 ⇒∀ 1≤k≤n→ (((2n+1)),(( 1)) )− (((2n+1)),(( 3)) ) tan^2 α_k +...=0 therefore “ x_k =tan^2 α_k ” thr roots of the equation are blowe x^n − (((2n+1)),((2n−1)) ) x^n + (((2n+1)),((2n−3)) )x^(n−1) −...=0 the sume of the roots of the equation is “ s= (((2n+1)),((2n−1)) ) ” ⇒s=Σ_(k=1) ^n tan^2 (((kπ)/(2n+1)))=n(2n+1)✓ the proof of the seconf part is done similarly

$${Answer}\:{to}\:{the}\:{question}\:``\mathrm{196008}'' \\ $$$$\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right) \\ $$$$\left.{Ans}\right) \\ $$$${according}\:\:``{de}\:{moivre}'' \\ $$$${sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}} {sin}\alpha−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\left({cos}\alpha\right)^{\mathrm{2}{n}−\mathrm{2}} \left({sin}\alpha\right)^{\mathrm{3}} +.... \\ $$$$=\left({cos}\alpha\right)^{\mathrm{2}{n}} \left({sin}\alpha\right)\left[\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha+...\right] \\ $$$${for}\:\:``\alpha_{{k}} =\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\:\:\:\:\:;\:\:\mathrm{1}\leqslant{k}\leqslant{n}\:\Rightarrow{sin}\left(\mathrm{2}{n}+\mathrm{1}\right)\alpha_{{k}} =\mathrm{0} \\ $$$$\Rightarrow\forall\:\:\mathrm{1}\leqslant{k}\leqslant{n}\rightarrow\:\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\mathrm{1}}\end{pmatrix}−\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\:\:\:\:\:\:\mathrm{3}}\end{pmatrix}\:{tan}^{\mathrm{2}} \alpha_{{k}} +...=\mathrm{0} \\ $$$${therefore}\:\:``\:{x}_{{k}} ={tan}^{\mathrm{2}} \alpha_{{k}} \:''\:{thr}\:{roots}\:{of}\:\:{the}\:{equation}\:{are}\:{blowe} \\ $$$${x}^{{n}} −\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:{x}^{{n}} +\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{3}}\end{pmatrix}{x}^{{n}−\mathrm{1}} −...=\mathrm{0} \\ $$$${the}\:{sume}\:{of}\:{the}\:{roots}\:{of}\:{the}\:{equation}\:{is}\:``\:{s}=\begin{pmatrix}{\mathrm{2}{n}+\mathrm{1}}\\{\mathrm{2}{n}−\mathrm{1}}\end{pmatrix}\:'' \\ $$$$\Rightarrow{s}=\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\:{tan}^{\mathrm{2}} \left(\frac{{k}\pi}{\mathrm{2}{n}+\mathrm{1}}\right)={n}\left(\mathrm{2}{n}+\mathrm{1}\right)\checkmark \\ $$$${the}\:{proof}\:{of}\:{the}\:{seconf}\:{part}\:{is}\:{done}\:{similarly} \\ $$$$ \\ $$$$ \\ $$

Question Number 196085    Answers: 2   Comments: 0

Question Number 196084    Answers: 1   Comments: 0

Question Number 196081    Answers: 0   Comments: 0

Question Number 196072    Answers: 2   Comments: 0

lim_(x→((3π)/2)) ((2x−3π−cot x)/(cos x+tan (((4x)/3))))

$$\:\: \\ $$$$ \underset{{x}\rightarrow\frac{\mathrm{3}\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{\mathrm{2}{x}−\mathrm{3}\pi−\mathrm{cot}\:{x}}{\mathrm{cos}\:{x}+\mathrm{tan}\:\left(\frac{\mathrm{4}{x}}{\mathrm{3}}\right)}\: \\ $$

Question Number 196070    Answers: 1   Comments: 0

During an invasion In the amazon dominion Every evening , every invader kills an amazon warrior Every morning , every amazon kills an invader soldier The 8^(th) day evening , there remains only one amazon and no invader How many were they in each part ?

$${During}\:{an}\:{invasion}\: \\ $$$${In}\:{the}\:{amazon}\:{dominion} \\ $$$${Every}\:{evening}\:,\:{every}\:\:{invader}\: \\ $$$${kills}\:\:{an}\:\:{amazon}\:{warrior}\: \\ $$$${Every}\:{morning}\:,\:{every}\:{amazon} \\ $$$${kills}\:\:\:{an}\:{invader}\:{soldier} \\ $$$${The}\:\mathrm{8}^{{th}} \:{day}\:{evening}\:,\:{there}\:{remains}\: \\ $$$${only}\:{one}\:{amazon}\:\:{and}\:{no}\:{invader} \\ $$$$ \\ $$$${How}\:{many}\:\:{were}\:{they}\:{in}\:{each}\:{part}\:? \\ $$

Question Number 196066    Answers: 2   Comments: 0

((( 1 −1)),((−1 2)) )^(−8) = ?

$$\begin{pmatrix}{\:\:\mathrm{1}\:\:\:\:\:\:\:\:\:−\mathrm{1}}\\{−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}}\end{pmatrix}^{−\mathrm{8}} =\:\:? \\ $$

Question Number 196063    Answers: 1   Comments: 0

calcul ∫_0 ^(+∞) (((√u) .arctan(u))/(1+u^2 )) du

$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\mathrm{calcul}\:\:\:\:\int_{\mathrm{0}} ^{+\infty} \:\frac{\sqrt{\mathrm{u}}\:.\mathrm{arctan}\left(\mathrm{u}\right)}{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\:\mathrm{du} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Question Number 196062    Answers: 1   Comments: 1

Σ_(i=1) ^n Σ_(j=1) ^n [gcd(i,j)=1]=? ,[D]= { ((1, D is ture.)),((0, D is false. )) :}

$$\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left[{gcd}\left({i},{j}\right)=\mathrm{1}\right]=?\:\:\:\:\:\:\:\:\:,\left[{D}\right]=\begin{cases}{\mathrm{1},\:\:\:{D}\:{is}\:{ture}.}\\{\mathrm{0},\:\:\:\:{D}\:{is}\:{false}.\:\:\:}\end{cases} \\ $$

Question Number 196056    Answers: 5   Comments: 0

solve (√(2x+3))−(√(3x+8))=(√(3x+4))−(√(2x+7))

$${solve} \\ $$$$\sqrt{\mathrm{2}{x}+\mathrm{3}}−\sqrt{\mathrm{3}{x}+\mathrm{8}}=\sqrt{\mathrm{3}{x}+\mathrm{4}}−\sqrt{\mathrm{2}{x}+\mathrm{7}} \\ $$

Question Number 196053    Answers: 1   Comments: 0

faind n terme 4,−2,((16)/9),−2,......

$${faind}\:{n}\:{terme} \\ $$$$\mathrm{4},−\mathrm{2},\frac{\mathrm{16}}{\mathrm{9}},−\mathrm{2},...... \\ $$

Question Number 196049    Answers: 1   Comments: 1

a, b, c > 0. Find the min value of Σ_(cyc) (√((a+b)/(a+b+c))) .

$${a},\:{b},\:{c}\:>\:\mathrm{0}.\:\mathrm{Find}\:\mathrm{the}\:\mathrm{min}\:\mathrm{value}\:\mathrm{of} \\ $$$$\underset{\mathrm{cyc}} {\sum}\:\sqrt{\frac{{a}+{b}}{{a}+{b}+{c}}}\:. \\ $$

Question Number 196046    Answers: 0   Comments: 0

∫e^x^2 ln^(24) (x)dx

$$\int{e}^{{x}^{\mathrm{2}} } \:{ln}^{\mathrm{24}} \left({x}\right){dx} \\ $$

Question Number 196037    Answers: 2   Comments: 0

∫ ((sin 2x)/(sin^3 x+cos^3 x)) dx =?

$$\:\:\:\int\:\frac{\mathrm{sin}\:\mathrm{2x}}{\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}}\:\mathrm{dx}\:=? \\ $$

Question Number 196035    Answers: 0   Comments: 0

Question Number 196024    Answers: 1   Comments: 0

lim_(x→0^+ ) [xΣ_(n=1) ^∞ ((1/n^(x+1) ))]=λ , evalute λ

$$\underset{{x}\rightarrow\mathrm{0}^{+} } {\mathrm{lim}}\left[{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{{n}^{{x}+\mathrm{1}} }\right)\right]=\lambda\:,\:{evalute}\:\lambda \\ $$

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