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Question Number 70886    Answers: 2   Comments: 11

Question Number 70847    Answers: 2   Comments: 0

Question Number 70842    Answers: 0   Comments: 1

Question Number 70838    Answers: 1   Comments: 0

Question Number 70834    Answers: 1   Comments: 0

Question Number 70831    Answers: 2   Comments: 1

Question Number 70826    Answers: 2   Comments: 1

Question Number 70818    Answers: 1   Comments: 1

what the prove that ∫_a ^b f(x) dx =∫_a ^b f(a+b−x) dx

$$\mathrm{what}\:\mathrm{the}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{x}\right)\:\mathrm{dx}\:=\int_{\mathrm{a}} ^{\mathrm{b}} \mathrm{f}\left(\mathrm{a}+\mathrm{b}−\mathrm{x}\right)\:\mathrm{dx} \\ $$

Question Number 70808    Answers: 1   Comments: 0

Question Number 70784    Answers: 1   Comments: 2

Question Number 70783    Answers: 1   Comments: 4

(5/(4∙9)) + 2((9/(16∙25))) + 3(((13)/(36∙49))) + 4(((17)/(64∙81))) + 5(((21)/(100∙121))) + … = ?

$$\frac{\mathrm{5}}{\mathrm{4}\centerdot\mathrm{9}}\:+\:\mathrm{2}\left(\frac{\mathrm{9}}{\mathrm{16}\centerdot\mathrm{25}}\right)\:+\:\mathrm{3}\left(\frac{\mathrm{13}}{\mathrm{36}\centerdot\mathrm{49}}\right)\:+\:\mathrm{4}\left(\frac{\mathrm{17}}{\mathrm{64}\centerdot\mathrm{81}}\right)\:+\:\mathrm{5}\left(\frac{\mathrm{21}}{\mathrm{100}\centerdot\mathrm{121}}\right)\:+\:\ldots\:=\:\:? \\ $$

Question Number 70780    Answers: 0   Comments: 1

(1/(t+(√u)+(√v)))= =(((t−(√u)+(√v))(t+(√u)−(√v))(t−(√u)−(√v)))/((t+(√u)+(√v))(t−(√u)+(√v))(t+(√u)−(√v))(t−(√u)−(√v))))= =((t^3 −((√u)+(√v))t^2 −((√u)−(√v))^2 t+(u−v)((√u)−(√v)))/(t^4 −2(u+v)t^2 +(u−v)^2 )) (1/(t+(u)^(1/3) +(v)^(1/3) ))= determinant (((α=−(1/2)+((√3)/2)i; β=−(1/2)−((√3)/2)i ⇒)),((⇒ α^2 =β; β^2 =α; α+β=−1; αβ=1))) =(((t+α(u)^(1/3) +β(v)^(1/3) )(t+β(u)^(1/3) +α(v)^(1/3) ))/((t+(u)^(1/3) +(v)^(1/3) )(t+α(u)^(1/3) +β(v)^(1/3) )(t+β(u)^(1/3) +α(v)^(1/3) )))= =((t^2 −((u)^(1/3) +(v)^(1/3) )t+(u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )/(t^3 +u+v−3((uv))^(1/3) t))= determinant (((a=t^3 +u+v; b=27uvt^3 )),(((1/(a−(b)^(1/3) ))=(((αa−β(b)^(1/3) )(βa−α(b)^(1/3) ))/((a−(b)^(1/3) )(αa−β(b)^(1/3) )(βa−α(b)^(1/3) )))=)),((=((a^2 +a(b)^(1/3) +(b^2 )^(1/3) )/(a^3 −b))))) =(N/(t^9 +3(u+v)t^6 +3(u^2 −7uv+v^2 )t^3 +(u+v)^3 )) N= =t^8 − −((u)^(1/3) +(v)^(1/3) )t^7 + +((u)^(1/3) +(v)^(1/3) )^2 t^6 + +((u)^(1/3) +(v)^(1/3) )((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^5 − −((u)^(1/3) +(v)^(1/3) )^2 ((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^4 + +((u)^(1/3) +(v)^(1/3) )^3 ((u)^(1/3) −2(v)^(1/3) )(2(u)^(1/3) −(v)^(1/3) )t^3 + +((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 t^2 − −((u)^(1/3) +(v)^(1/3) )((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 t+ +((u)^(1/3) +(v)^(1/3) )^2 ((u^2 )^(1/3) −((uv))^(1/3) +(v^2 )^(1/3) )^3 I think there′s no easier way...

$$\frac{\mathrm{1}}{{t}+\sqrt{{u}}+\sqrt{{v}}}= \\ $$$$=\frac{\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}{\left({t}+\sqrt{{u}}+\sqrt{{v}}\right)\left({t}−\sqrt{{u}}+\sqrt{{v}}\right)\left({t}+\sqrt{{u}}−\sqrt{{v}}\right)\left({t}−\sqrt{{u}}−\sqrt{{v}}\right)}= \\ $$$$=\frac{{t}^{\mathrm{3}} −\left(\sqrt{{u}}+\sqrt{{v}}\right){t}^{\mathrm{2}} −\left(\sqrt{{u}}−\sqrt{{v}}\right)^{\mathrm{2}} {t}+\left({u}−{v}\right)\left(\sqrt{{u}}−\sqrt{{v}}\right)}{{t}^{\mathrm{4}} −\mathrm{2}\left({u}+{v}\right){t}^{\mathrm{2}} +\left({u}−{v}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\frac{\mathrm{1}}{{t}+\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}}= \\ $$$$\:\:\:\:\:\begin{vmatrix}{\alpha=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i};\:\beta=−\frac{\mathrm{1}}{\mathrm{2}}−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{i}\:\Rightarrow}\\{\Rightarrow\:\alpha^{\mathrm{2}} =\beta;\:\beta^{\mathrm{2}} =\alpha;\:\alpha+\beta=−\mathrm{1};\:\alpha\beta=\mathrm{1}}\end{vmatrix} \\ $$$$=\frac{\left({t}+\alpha\sqrt[{\mathrm{3}}]{{u}}+\beta\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\beta\sqrt[{\mathrm{3}}]{{u}}+\alpha\sqrt[{\mathrm{3}}]{{v}}\right)}{\left({t}+\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\alpha\sqrt[{\mathrm{3}}]{{u}}+\beta\sqrt[{\mathrm{3}}]{{v}}\right)\left({t}+\beta\sqrt[{\mathrm{3}}]{{u}}+\alpha\sqrt[{\mathrm{3}}]{{v}}\right)}= \\ $$$$=\frac{{t}^{\mathrm{2}} −\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right){t}+\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }}{{t}^{\mathrm{3}} +{u}+{v}−\mathrm{3}\sqrt[{\mathrm{3}}]{{uv}}\:{t}}= \\ $$$$\:\:\:\:\:\begin{vmatrix}{{a}={t}^{\mathrm{3}} +{u}+{v};\:{b}=\mathrm{27}{uvt}^{\mathrm{3}} }\\{\frac{\mathrm{1}}{{a}−\sqrt[{\mathrm{3}}]{{b}}}=\frac{\left(\alpha{a}−\beta\sqrt[{\mathrm{3}}]{{b}}\right)\left(\beta{a}−\alpha\sqrt[{\mathrm{3}}]{{b}}\right)}{\left({a}−\sqrt[{\mathrm{3}}]{{b}}\right)\left(\alpha{a}−\beta\sqrt[{\mathrm{3}}]{{b}}\right)\left(\beta{a}−\alpha\sqrt[{\mathrm{3}}]{{b}}\right)}=}\\{=\frac{{a}^{\mathrm{2}} +{a}\sqrt[{\mathrm{3}}]{{b}}+\sqrt[{\mathrm{3}}]{{b}^{\mathrm{2}} }}{{a}^{\mathrm{3}} −{b}}}\end{vmatrix} \\ $$$$=\frac{{N}}{{t}^{\mathrm{9}} +\mathrm{3}\left({u}+{v}\right){t}^{\mathrm{6}} +\mathrm{3}\left({u}^{\mathrm{2}} −\mathrm{7}{uv}+{v}^{\mathrm{2}} \right){t}^{\mathrm{3}} +\left({u}+{v}\right)^{\mathrm{3}} } \\ $$$${N}= \\ $$$$={t}^{\mathrm{8}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{7}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} {t}^{\mathrm{6}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{5}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{4}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{3}} \left(\sqrt[{\mathrm{3}}]{{u}}−\mathrm{2}\sqrt[{\mathrm{3}}]{{v}}\right)\left(\mathrm{2}\sqrt[{\mathrm{3}}]{{u}}−\sqrt[{\mathrm{3}}]{{v}}\right){t}^{\mathrm{3}} + \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} {t}^{\mathrm{2}} − \\ $$$$−\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)\left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} {t}+ \\ $$$$+\left(\sqrt[{\mathrm{3}}]{{u}}+\sqrt[{\mathrm{3}}]{{v}}\right)^{\mathrm{2}} \left(\sqrt[{\mathrm{3}}]{{u}^{\mathrm{2}} }−\sqrt[{\mathrm{3}}]{{uv}}+\sqrt[{\mathrm{3}}]{{v}^{\mathrm{2}} }\right)^{\mathrm{3}} \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{think}\:\mathrm{there}'\mathrm{s}\:\mathrm{no}\:\mathrm{easier}\:\mathrm{way}... \\ $$

Question Number 70768    Answers: 1   Comments: 1

Question Number 70757    Answers: 0   Comments: 3

.

$$. \\ $$

Question Number 70756    Answers: 0   Comments: 1

I cannot understand all those who post questions like ∫x^(Γ(x^2 )) cos ((log_(ϖ+x) x^(2πi) ))^(1/x) dx=? and a few minutes later (5/3)×((2+1)/(9−4))=? I mean, are you serious?

$$\mathrm{I}\:\mathrm{cannot}\:\mathrm{understand}\:\mathrm{all}\:\mathrm{those}\:\mathrm{who}\:\mathrm{post} \\ $$$$\mathrm{questions}\:\mathrm{like}\:\int{x}^{\Gamma\left({x}^{\mathrm{2}} \right)} \mathrm{cos}\:\sqrt[{{x}}]{\mathrm{log}_{\varpi+{x}} \:{x}^{\mathrm{2}\pi\mathrm{i}} }{dx}=? \\ $$$$\mathrm{and}\:\mathrm{a}\:\mathrm{few}\:\mathrm{minutes}\:\mathrm{later}\:\frac{\mathrm{5}}{\mathrm{3}}×\frac{\mathrm{2}+\mathrm{1}}{\mathrm{9}−\mathrm{4}}=? \\ $$$$\mathrm{I}\:\mathrm{mean},\:\mathrm{are}\:\mathrm{you}\:\mathrm{serious}? \\ $$

Question Number 70782    Answers: 0   Comments: 3

how can i get my password ? i can′t remeber

$$\mathrm{how}\:\mathrm{can}\:\mathrm{i}\:\mathrm{get}\:\mathrm{my}\:\mathrm{password}\:?\:\mathrm{i}\:\:\mathrm{can}'\mathrm{t}\:\mathrm{remeber} \\ $$

Question Number 70746    Answers: 0   Comments: 4

Question Number 70742    Answers: 0   Comments: 1

Prove that f(x) = x − a cos (x) − b has at least one real root for ∀a,b ∈ R

$$\mathrm{Prove}\:\mathrm{that}\:{f}\left({x}\right)\:=\:{x}\:−\:{a}\:\mathrm{cos}\:\left({x}\right)\:−\:{b} \\ $$$$\mathrm{has}\:\mathrm{at}\:\mathrm{least}\:\mathrm{one}\:\mathrm{real}\:\mathrm{root}\:\mathrm{for}\:\forall{a},{b}\:\in\:\mathbb{R} \\ $$

Question Number 70738    Answers: 1   Comments: 2

Question Number 70741    Answers: 0   Comments: 5

f(x)= { ((x^2 x≤1)),(),((∣x−2∣ x>1)) :} find the critical points

$${f}\left({x}\right)=\begin{cases}{{x}^{\mathrm{2}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\leqslant\mathrm{1}}\\{}\\{\mid{x}−\mathrm{2}\mid\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}>\mathrm{1}}\end{cases} \\ $$$$ \\ $$$${find}\:{the}\:{critical}\:{points} \\ $$

Question Number 70798    Answers: 1   Comments: 1

Question Number 70720    Answers: 2   Comments: 4

y′′+((y′)/x)+((4y)/x^2 )=1+x^2

$$\mathrm{y}''+\frac{\mathrm{y}'}{\mathrm{x}}+\frac{\mathrm{4y}}{\mathrm{x}^{\mathrm{2}} }=\mathrm{1}+\mathrm{x}^{\mathrm{2}} \\ $$

Question Number 70710    Answers: 1   Comments: 0

pleace how can link this app to my telegram

$$\mathrm{pleace}\:\mathrm{how}\:\mathrm{can}\:\mathrm{link}\:\mathrm{this}\: \\ $$$$\mathrm{app}\:\mathrm{to}\:\mathrm{my}\:\mathrm{telegram} \\ $$

Question Number 70713    Answers: 0   Comments: 1

can i login to my account with other phone? if so how

$$\mathrm{can}\:\mathrm{i}\:\mathrm{login}\:\mathrm{to}\:\mathrm{my}\:\mathrm{account}\: \\ $$$$\mathrm{with}\:\mathrm{other}\:\mathrm{phone}?\:\mathrm{if}\:\mathrm{so}\:\mathrm{how} \\ $$

Question Number 70705    Answers: 1   Comments: 0

(d/dx)(log_(10) x)=?

$$\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{log}_{\mathrm{10}} \mathrm{x}\right)=? \\ $$

Question Number 70704    Answers: 1   Comments: 2

z^4 −2z^2 +2=0

$$\mathrm{z}^{\mathrm{4}} −\mathrm{2z}^{\mathrm{2}} +\mathrm{2}=\mathrm{0} \\ $$

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