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Question Number 156709    Answers: 1   Comments: 0

solve (x^3 )^(1/(1/x)) = 27

$$\boldsymbol{{solve}}\:\sqrt[{\frac{\mathrm{1}}{\boldsymbol{{x}}}}]{\boldsymbol{{x}}^{\mathrm{3}} }\:=\:\mathrm{27} \\ $$

Question Number 156695    Answers: 1   Comments: 0

∫sin(ln(x))dx=?

$$\int{sin}\left({ln}\left({x}\right)\right){dx}=? \\ $$

Question Number 156691    Answers: 1   Comments: 0

Show that (Σ_(k=1) ^n x_k y_k )^2 ≤(Σ_(k=1) ^n x_k ^2 )×(Σ_(k=1) ^n y_k ^2 )t

$${Show}\:{that} \\ $$$$\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{k}} {y}_{{k}} \right)^{\mathrm{2}} \leqslant\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{x}_{{k}} ^{\mathrm{2}} \right)×\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{y}_{{k}} ^{\mathrm{2}} \right){t} \\ $$

Question Number 156689    Answers: 0   Comments: 0

Question Number 156684    Answers: 2   Comments: 0

solve the D.E [1+e^(x/y) ]dx+e^(x/y) [1−(x/y)]dy=0 any one can help pls

$${solve}\:{the}\:{D}.{E}\: \\ $$$$\left[\mathrm{1}+{e}^{\frac{{x}}{{y}}} \right]{dx}+{e}^{\frac{{x}}{{y}}} \left[\mathrm{1}−\frac{{x}}{{y}}\right]{dy}=\mathrm{0} \\ $$$${any}\:{one}\:{can}\:{help}\:{pls} \\ $$

Question Number 156683    Answers: 1   Comments: 3

Question Number 156677    Answers: 1   Comments: 3

Question Number 156676    Answers: 3   Comments: 1

Question Number 156671    Answers: 2   Comments: 0

Question Number 156663    Answers: 0   Comments: 2

calcul ∫_0 ^(10) e^(x−E(x)) dx (E la partie entiere) besoin d′aide

$${calcul}\:\int_{\mathrm{0}} ^{\mathrm{10}} {e}^{{x}−{E}\left({x}\right)} {dx}\:\left({E}\:{la}\:{partie}\:{entiere}\right) \\ $$$${besoin}\:{d}'{aide} \\ $$

Question Number 156662    Answers: 0   Comments: 0

I =∫_( 0) ^( 1) ((2 ln x - x + (1/2))/((1 + x^2 ) ln^3 x)) dx = ?

$$\boldsymbol{\mathrm{I}}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{2}\:\mathrm{ln}\:\mathrm{x}\:-\:\mathrm{x}\:+\:\frac{\mathrm{1}}{\mathrm{2}}}{\left(\mathrm{1}\:+\:\mathrm{x}^{\mathrm{2}} \right)\:\mathrm{ln}^{\mathrm{3}} \:\mathrm{x}}\:\mathrm{dx}\:=\:? \\ $$

Question Number 156661    Answers: 0   Comments: 0

if 1≤x ; y≤2 ; 3≤z ; t≤4 prove that: ((x^2 +y^2 +1)/(xz+yt+3)) + ((√6)/3) ≤ ((3+xz+yt)/(9+z^2 +t^2 )) + ((11)/(12))

$$\mathrm{if}\:\:\mathrm{1}\leqslant\mathrm{x}\:\:;\:\:\mathrm{y}\leqslant\mathrm{2}\:\:;\:\:\mathrm{3}\leqslant\mathrm{z}\:\:;\:\:\mathrm{t}\leqslant\mathrm{4}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{1}}{\mathrm{xz}+\mathrm{yt}+\mathrm{3}}\:+\:\frac{\sqrt{\mathrm{6}}}{\mathrm{3}}\:\leqslant\:\frac{\mathrm{3}+\mathrm{xz}+\mathrm{yt}}{\mathrm{9}+\mathrm{z}^{\mathrm{2}} +\mathrm{t}^{\mathrm{2}} }\:+\:\frac{\mathrm{11}}{\mathrm{12}} \\ $$

Question Number 156660    Answers: 0   Comments: 0

Question Number 156652    Answers: 1   Comments: 0

Sirs, please give me the general solutions to a quadratic ineqality. (1) ax^2 + bx + c > 0 (2) ax^2 + bx + c ≥ 0 (3) ax^2 + bx + c < 0 (4) ax^2 + bx + c ≤ 0

$$\mathrm{Sirs},\:\mathrm{please}\:\mathrm{give}\:\mathrm{me}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{a}\:\mathrm{quadratic}\:\mathrm{ineqality}. \\ $$$$\left(\mathrm{1}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:>\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{2}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\geqslant\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:<\:\:\:\:\mathrm{0} \\ $$$$\left(\mathrm{4}\right)\:\:\:\:\:\:\mathrm{ax}^{\mathrm{2}} \:\:\:+\:\:\:\mathrm{bx}\:\:\:+\:\:\:\mathrm{c}\:\:\:\:\leqslant\:\:\:\:\mathrm{0} \\ $$

Question Number 156648    Answers: 3   Comments: 0

Solve for integers: x^2 y^2 - x^2 - xy - y^2 + x + y = 0

$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{integers}: \\ $$$$\mathrm{x}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} \:-\:\mathrm{x}^{\mathrm{2}} \:-\:\mathrm{xy}\:-\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{x}\:+\:\mathrm{y}\:=\:\mathrm{0} \\ $$

Question Number 156647    Answers: 0   Comments: 0

Question Number 156642    Answers: 1   Comments: 3

For x∈(0;∞) - Z prove that: (({x}^3 )/([x])) + (([x]^3 )/({x})) ≥ (1/8) (x^2 + [x]^2 + {x}^2 ) [∗]-GIF and {x}=x-[x]

$$\mathrm{For}\:\:\mathrm{x}\in\left(\mathrm{0};\infty\right)\:-\:\mathbb{Z}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\frac{\left\{\mathrm{x}\right\}^{\mathrm{3}} }{\left[\mathrm{x}\right]}\:+\:\frac{\left[\mathrm{x}\right]^{\mathrm{3}} }{\left\{\mathrm{x}\right\}}\:\geqslant\:\frac{\mathrm{1}}{\mathrm{8}}\:\left(\mathrm{x}^{\mathrm{2}} \:+\:\left[\mathrm{x}\right]^{\mathrm{2}} \:+\:\left\{\mathrm{x}\right\}^{\mathrm{2}} \right) \\ $$$$\left[\ast\right]-\mathrm{GIF}\:\:\mathrm{and}\:\:\left\{\mathrm{x}\right\}=\mathrm{x}-\left[\mathrm{x}\right] \\ $$

Question Number 156640    Answers: 1   Comments: 0

Question Number 156639    Answers: 0   Comments: 0

x^3 =x+c x^4 =x^2 +cx let cx=kx^4 +hx^2 ⇒ kx^3 +hx=c x^2 =1+c(kx^2 +h) x^2 =((1+ch)/(1−ck)) (((1+ch)/(1−ck))){((k(1+ch))/(1−ck))+h}^2 =c^2 (1+ch)(h+k)^2 =c^2 (1−ck)^3 ⇒ (1+ch)(h^2 +k^2 +2hk) =c^2 (1−c^3 k^3 +3c^2 k^2 −3ck) c^5 k^3 +(1+ch−3c^4 )k^2 +{2h(1+ch)+3c^3 }k +(1+ch)h^2 −c^2 =0 let h=3c^3 −(1/c) ⇒ c^5 k^3 +{9c^3 −(2/c)+2c(3c^3 −(1/c))^2 }k +{3c^4 (3c^3 −(1/c))^2 −c^2 }=0 ⇒ k^3 +3(6c^2 −(1/c^2 ))k +27c^5 −18c+(2/c^3 )=0 D=(((27c^5 )/2)+9c+(1/c^3 ))^2 +(6c^2 −(1/c^2 ))^3 ...

$$\mathrm{x}^{\mathrm{3}} =\mathrm{x}+\mathrm{c} \\ $$$$\mathrm{x}^{\mathrm{4}} =\mathrm{x}^{\mathrm{2}} +\mathrm{cx} \\ $$$$\mathrm{let}\:\:\mathrm{cx}=\mathrm{kx}^{\mathrm{4}} +\mathrm{hx}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{kx}^{\mathrm{3}} +\mathrm{hx}=\mathrm{c} \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{1}+\mathrm{c}\left(\mathrm{kx}^{\mathrm{2}} +\mathrm{h}\right) \\ $$$$\mathrm{x}^{\mathrm{2}} =\frac{\mathrm{1}+\mathrm{ch}}{\mathrm{1}−\mathrm{ck}} \\ $$$$\left(\frac{\mathrm{1}+\mathrm{ch}}{\mathrm{1}−\mathrm{ck}}\right)\left\{\frac{\mathrm{k}\left(\mathrm{1}+\mathrm{ch}\right)}{\mathrm{1}−\mathrm{ck}}+\mathrm{h}\right\}^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} \\ $$$$\left(\mathrm{1}+\mathrm{ch}\right)\left(\mathrm{h}+\mathrm{k}\right)^{\mathrm{2}} =\mathrm{c}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{ck}\right)^{\mathrm{3}} \\ $$$$\Rightarrow\:\:\left(\mathrm{1}+\mathrm{ch}\right)\left(\mathrm{h}^{\mathrm{2}} +\mathrm{k}^{\mathrm{2}} +\mathrm{2hk}\right) \\ $$$$\:\:\:\:=\mathrm{c}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{c}^{\mathrm{3}} \mathrm{k}^{\mathrm{3}} +\mathrm{3c}^{\mathrm{2}} \mathrm{k}^{\mathrm{2}} −\mathrm{3ck}\right) \\ $$$$\mathrm{c}^{\mathrm{5}} \mathrm{k}^{\mathrm{3}} +\left(\mathrm{1}+\mathrm{ch}−\mathrm{3c}^{\mathrm{4}} \right)\mathrm{k}^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\left\{\mathrm{2h}\left(\mathrm{1}+\mathrm{ch}\right)+\mathrm{3c}^{\mathrm{3}} \right\}\mathrm{k} \\ $$$$\:\:\:\:\:+\left(\mathrm{1}+\mathrm{ch}\right)\mathrm{h}^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{let}\:\:\mathrm{h}=\mathrm{3c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{c}}\:\:\Rightarrow \\ $$$$\mathrm{c}^{\mathrm{5}} \mathrm{k}^{\mathrm{3}} +\left\{\mathrm{9c}^{\mathrm{3}} −\frac{\mathrm{2}}{\mathrm{c}}+\mathrm{2c}\left(\mathrm{3c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{c}}\right)^{\mathrm{2}} \right\}\mathrm{k} \\ $$$$\:\:\:\:+\left\{\mathrm{3c}^{\mathrm{4}} \left(\mathrm{3c}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{c}}\right)^{\mathrm{2}} −\mathrm{c}^{\mathrm{2}} \right\}=\mathrm{0} \\ $$$$\Rightarrow\:\mathrm{k}^{\mathrm{3}} +\mathrm{3}\left(\mathrm{6c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\right)\mathrm{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\mathrm{27c}^{\mathrm{5}} −\mathrm{18c}+\frac{\mathrm{2}}{\mathrm{c}^{\mathrm{3}} }=\mathrm{0} \\ $$$$\mathrm{D}=\left(\frac{\mathrm{27c}^{\mathrm{5}} }{\mathrm{2}}+\mathrm{9c}+\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{3}} }\right)^{\mathrm{2}} +\left(\mathrm{6c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{c}^{\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$$... \\ $$

Question Number 156617    Answers: 1   Comments: 0

Question Number 156616    Answers: 2   Comments: 0

∫_0 ^1 ((arcsin(x))/x)dx=?

$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{arcsin}\left({x}\right)}{{x}}{dx}=? \\ $$

Question Number 156636    Answers: 1   Comments: 0

Question Number 156611    Answers: 0   Comments: 1

lim(√(x+5))

$${lim}\sqrt{{x}+\mathrm{5}} \\ $$

Question Number 156610    Answers: 2   Comments: 0

If A and B are invertible matrices,then: (AB)^(−1) =B^(−1) A^(−1) ≠ A^(−1) B^(−1) proove.

$$\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{invertible}\:\mathrm{matrices},\mathrm{then}: \\ $$$$\left(\mathrm{AB}\right)^{−\mathrm{1}} =\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \neq\:\mathrm{A}^{−\mathrm{1}} \mathrm{B}^{−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{proove}}. \\ $$

Question Number 156609    Answers: 1   Comments: 0

Question Number 156604    Answers: 1   Comments: 2

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