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Question Number 164425    Answers: 1   Comments: 0

What it′s ?? 5×55×555×5555×55555.............∞=

$$\mathrm{What}\:\mathrm{it}'\mathrm{s}\:?? \\ $$$$\mathrm{5}×\mathrm{55}×\mathrm{555}×\mathrm{5555}×\mathrm{55555}.............\infty= \\ $$

Question Number 164419    Answers: 1   Comments: 0

please help me prouve that ∫_0 ^1 ((lnt)/(t^2 −1))dt=(π^2 /8)

$${please}\:{help}\:{me} \\ $$$${prouve}\:{that}\:\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{lnt}}{{t}^{\mathrm{2}} −\mathrm{1}}{dt}=\frac{\pi^{\mathrm{2}} }{\mathrm{8}} \\ $$

Question Number 164418    Answers: 2   Comments: 2

Question Number 164416    Answers: 0   Comments: 0

if a;b;c<0 and a+b+c=3 prove that: (1 + (b/a))^(1/b^2 ) ∙ (1 + (c/b))^(1/c^2 ) ∙ (1 + (a/c))^(1/a^2 ) ≥ 8

$$\mathrm{if}\:\:\mathrm{a};\mathrm{b};\mathrm{c}<\mathrm{0}\:\:\mathrm{and}\:\:\mathrm{a}+\mathrm{b}+\mathrm{c}=\mathrm{3}\:\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\left(\mathrm{1}\:+\:\frac{\mathrm{b}}{\mathrm{a}}\right)^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }} \centerdot\:\left(\mathrm{1}\:+\:\frac{\mathrm{c}}{\mathrm{b}}\right)^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{c}}^{\mathrm{2}} }} \centerdot\:\left(\mathrm{1}\:+\:\frac{\mathrm{a}}{\mathrm{c}}\right)^{\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} }} \geqslant\:\mathrm{8} \\ $$

Question Number 164417    Answers: 1   Comments: 0

Question Number 164405    Answers: 0   Comments: 0

let a, b, c > 0 ; a + b + c = 3 prove that; (a^2 /(b^2 + bc + c^2 )) + (b^2 /(c^2 + ac + a^2 )) + (c^2 /(b^2 +ba + a^2 )) 6 ≥ 2abc + (5/3) (a^2 + b^2 + c^2 ) ≥ 7 ^({Z.A})

$$\boldsymbol{{let}}\:\boldsymbol{{a}},\:\boldsymbol{{b}},\:\boldsymbol{{c}}\:>\:\mathrm{0}\:;\:\boldsymbol{{a}}\:+\:\boldsymbol{{b}}\:+\:\boldsymbol{{c}}\:=\:\mathrm{3} \\ $$$$\boldsymbol{{prove}}\:\boldsymbol{{that}};\:\frac{\boldsymbol{{a}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} \:+\:\boldsymbol{{bc}}\:+\:\boldsymbol{{c}}^{\mathrm{2}} }\:\:+\:\:\frac{\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{c}}^{\mathrm{2}} \:+\:\boldsymbol{{ac}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} \:}\:\:+\:\:\frac{\boldsymbol{{c}}^{\mathrm{2}} }{\boldsymbol{{b}}^{\mathrm{2}} \:+\boldsymbol{{ba}}\:+\:\boldsymbol{{a}}^{\mathrm{2}} }\:\:\mathrm{6}\:\:\geqslant\:\:\mathrm{2}\boldsymbol{{abc}}\:\:+\:\:\frac{\mathrm{5}}{\mathrm{3}}\:\:\left(\boldsymbol{{a}}^{\mathrm{2}} \:+\:\boldsymbol{{b}}^{\mathrm{2}} \:+\:\boldsymbol{{c}}^{\mathrm{2}} \:\right)\:\:\geqslant\:\:\mathrm{7} \\ $$$$\:^{\left\{\boldsymbol{\mathrm{Z}}.\mathrm{A}\right\}} \\ $$

Question Number 164396    Answers: 2   Comments: 1

If f(x)=f(x−1)+x^2 +2x and f(0)=17 , find f(17).

$${If}\:\:{f}\left({x}\right)={f}\left({x}−\mathrm{1}\right)+{x}^{\mathrm{2}} +\mathrm{2}{x} \\ $$$${and}\:\:{f}\left(\mathrm{0}\right)=\mathrm{17}\:\:,\:\:{find}\:{f}\left(\mathrm{17}\right). \\ $$

Question Number 164395    Answers: 2   Comments: 0

Question Number 164391    Answers: 1   Comments: 0

If { ((sin (A−B)=(1/( (√(10)))))),((cos (A+B)=(2/( (√(29)))))) :}; 0<A<(π/4) ; 0<B<(π/4) Find tan 2A.

$$\:\:\mathrm{If}\:\begin{cases}{\mathrm{sin}\:\left(\mathrm{A}−\mathrm{B}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{10}}}}\\{\mathrm{cos}\:\left(\mathrm{A}+\mathrm{B}\right)=\frac{\mathrm{2}}{\:\sqrt{\mathrm{29}}}}\end{cases};\:\mathrm{0}<\mathrm{A}<\frac{\pi}{\mathrm{4}}\:;\:\mathrm{0}<\mathrm{B}<\frac{\pi}{\mathrm{4}} \\ $$$$\:\mathrm{Find}\:\mathrm{tan}\:\mathrm{2A}. \\ $$

Question Number 164386    Answers: 0   Comments: 8

CH_3 −CH_∣_(CH_3 ) −CH=C^∣^(C^(∣CH_3 ) H_2 ) −CH_∣_(CH_∥_(CH_2 ) ) −CH_∣_Δ −CH_∣_(Br) −CH_3 what is the iupac name?

$${CH}_{\mathrm{3}} −{C}\underset{\underset{{CH}_{\mathrm{3}} } {\mid}} {{H}}−{CH}=\overset{\overset{\overset{\mid{CH}_{\mathrm{3}} } {{C}H}_{\mathrm{2}} } {\mid}} {{C}}−{C}\underset{\underset{{C}\underset{\underset{{CH}_{\mathrm{2}} } {\parallel}} {{H}}} {\mid}} {{H}}−{C}\underset{\underset{\Delta} {\mid}} {{H}}−{C}\underset{\underset{{Br}} {\mid}} {{H}}−{CH}_{\mathrm{3}} \:\:\:\: \\ $$$${what}\:{is}\:{the}\:{iupac}\:{name}? \\ $$

Question Number 164496    Answers: 2   Comments: 0

{ ((x^2 + xy = 11)),((y^2 + xy = 24)) :} ⇒ ∣x+y∣=?

$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{11}}\\{\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{xy}\:=\:\mathrm{24}}\end{cases}\:\:\Rightarrow\:\mid\mathrm{x}+\mathrm{y}\mid=? \\ $$

Question Number 164376    Answers: 2   Comments: 0

Question Number 164374    Answers: 0   Comments: 2

Question Number 164372    Answers: 1   Comments: 1

Question Number 164367    Answers: 1   Comments: 0

∫ e^(tan(x)) dx {Z.A}

$$\int\:\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \:\boldsymbol{{dx}} \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{{A}}\right\} \\ $$

Question Number 164366    Answers: 1   Comments: 1

(d/dx) (e^(tan(x)) ) {Z.A}

$$\frac{\boldsymbol{{d}}}{\boldsymbol{{dx}}}\:\left(\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \right) \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{\mathrm{A}}\right\} \\ $$

Question Number 164364    Answers: 0   Comments: 0

Lim_(x→∞) (e^(tan(x)) ) {Z.A}

$$\boldsymbol{{Lim}}_{\boldsymbol{{x}}\rightarrow\infty} \:\left(\boldsymbol{{e}}^{\boldsymbol{{tan}}\left(\boldsymbol{{x}}\right)} \right) \\ $$$$\left\{\boldsymbol{{Z}}.\boldsymbol{\mathrm{A}}\right\} \\ $$

Question Number 164347    Answers: 0   Comments: 4

Question Number 164345    Answers: 2   Comments: 1

Question Number 164341    Answers: 0   Comments: 1

x^4 =x^2 +cx (x^2 −(1/2))^2 =cx+(1/4) let x^2 −(1/2)=t ⇒ (t^2 −(1/4))^2 =c^2 (t+(1/2)) now let t^2 −(1/4)=z ⇒ (z^2 −(c^2 /2))^2 =c^4 (z+(1/4)) z^2 −(c^2 /2)=p ⇒ (p^2 −(c^4 /4))^2 =c^8 (p+(c^2 /2)) ⇒ p^4 −((c^4 p^2 )/2)−c^8 p+(c^8 /(16))−(c^(10) /2)=0 p^4 −Ap^2 −Bp−λ=0 (p^2 +sp+h)(p^2 −sp−(λ/h))=0 h−(λ/h)=s^2 −A s(h+(λ/h))=B (B^2 /s^2 )−(s^2 −A)^2 =4λ (s^2 −A)^3 +A(s^2 −A)^2 +4λ(s^2 −A) +4λA−B^2 =0 m^3 +Am^2 +4λm+(4λA−B^2 )=0 let m=w−(A/3) ⇒ w^3 +(4λ−(A^2 /3))w+((2A^3 )/(27))+((8λA)/3)−B^2 =0 4λ−(A^2 /3)=2c^(10) −(c^8 /4)−(c^8 /(12)) =2c^8 (c^2 −(1/6))

$$\:\:{x}^{\mathrm{4}} ={x}^{\mathrm{2}} +{cx} \\ $$$$\left({x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ={cx}+\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${let}\:\:\:\:{x}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}={t}\:\:\:\Rightarrow \\ $$$$\left({t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{2}} \left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${now}\:\:{let}\:\:\:{t}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}={z}\:\:\:\Rightarrow \\ $$$$\left({z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right)^{\mathrm{2}} ={c}^{\mathrm{4}} \left({z}+\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$${z}^{\mathrm{2}} −\frac{{c}^{\mathrm{2}} }{\mathrm{2}}={p}\:\:\Rightarrow\:\:\left({p}^{\mathrm{2}} −\frac{{c}^{\mathrm{4}} }{\mathrm{4}}\right)^{\mathrm{2}} ={c}^{\mathrm{8}} \left({p}+\frac{{c}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\:{p}^{\mathrm{4}} −\frac{{c}^{\mathrm{4}} {p}^{\mathrm{2}} }{\mathrm{2}}−{c}^{\mathrm{8}} {p}+\frac{{c}^{\mathrm{8}} }{\mathrm{16}}−\frac{{c}^{\mathrm{10}} }{\mathrm{2}}=\mathrm{0} \\ $$$${p}^{\mathrm{4}} −{Ap}^{\mathrm{2}} −{Bp}−\lambda=\mathrm{0} \\ $$$$\left({p}^{\mathrm{2}} +{sp}+{h}\right)\left({p}^{\mathrm{2}} −{sp}−\frac{\lambda}{{h}}\right)=\mathrm{0} \\ $$$${h}−\frac{\lambda}{{h}}={s}^{\mathrm{2}} −{A} \\ $$$${s}\left({h}+\frac{\lambda}{{h}}\right)={B} \\ $$$$\frac{{B}^{\mathrm{2}} }{{s}^{\mathrm{2}} }−\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} =\mathrm{4}\lambda \\ $$$$\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{3}} +{A}\left({s}^{\mathrm{2}} −{A}\right)^{\mathrm{2}} +\mathrm{4}\lambda\left({s}^{\mathrm{2}} −{A}\right) \\ $$$$\:\:\:\:\:\:\:\:\:+\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$${m}^{\mathrm{3}} +{Am}^{\mathrm{2}} +\mathrm{4}\lambda{m}+\left(\mathrm{4}\lambda{A}−{B}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${let}\:\:\:{m}={w}−\frac{{A}}{\mathrm{3}}\:\:\:\Rightarrow \\ $$$${w}^{\mathrm{3}} +\left(\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}\right){w}+\frac{\mathrm{2}{A}^{\mathrm{3}} }{\mathrm{27}}+\frac{\mathrm{8}\lambda{A}}{\mathrm{3}}−{B}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{4}\lambda−\frac{{A}^{\mathrm{2}} }{\mathrm{3}}=\mathrm{2}{c}^{\mathrm{10}} −\frac{{c}^{\mathrm{8}} }{\mathrm{4}}−\frac{{c}^{\mathrm{8}} }{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}{c}^{\mathrm{8}} \left({c}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$

Question Number 164339    Answers: 1   Comments: 0

In AB^Δ C : cos^( 2) (A )+ cos^( 2) (B )+ cos^( 2) ( C )=1 . Prove that AB^Δ C is right angled. −−−−−−−−

$$ \\ $$$$\:\:\:\:{In}\:\:{A}\overset{\Delta} {{B}C}\:\:\::\:\:\:{cos}^{\:\mathrm{2}} \left({A}\:\right)+\:{cos}^{\:\mathrm{2}} \left({B}\:\right)+\:{cos}^{\:\mathrm{2}} \left(\:{C}\:\right)=\mathrm{1}\:\:. \\ $$$$\:\:\:\:\:\:\:\:{Prove}\:{that}\:\:{A}\overset{\Delta} {{B}C}\:\:\:{is}\:\:\:{right}\:{angled}. \\ $$$$\:\:\:\:\:\:−−−−−−−− \\ $$$$\:\:\:\:\:\: \\ $$

Question Number 164331    Answers: 0   Comments: 2

Question Number 164328    Answers: 2   Comments: 0

solve Σ_(n=0) ^∞ (( cos^(2n) (x))/(n!)) = (e)^(1/4) ■ −−−−−−−

$$ \\ $$$$\:\:\:\:\:{solve} \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:{cos}^{\mathrm{2}{n}} \left({x}\right)}{{n}!}\:=\:\sqrt[{\mathrm{4}}]{{e}}\:\:\:\:\:\:\:\:\:\:\:\blacksquare\: \\ $$$$\:\:\:\:\:\:\:\:−−−−−−− \\ $$$$\:\: \\ $$

Question Number 164315    Answers: 0   Comments: 3

Question Number 164312    Answers: 0   Comments: 3

Question Number 164311    Answers: 0   Comments: 0

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