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Q1 ∴ x=<1a_1 a_2 ...a_n >∈N & y=<a_1 a_2 ...a_n 1>∈N if y=3x then , find the smallest value of x Q2 ∴ with the above conditions ,what other values can be placed besides the number “ 1 ”

$${Q}\mathrm{1}\:\therefore\:\:{x}=<\mathrm{1}{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} >\in\mathbb{N}\:\:\&\:\:{y}=<{a}_{\mathrm{1}} {a}_{\mathrm{2}} ...{a}_{{n}} \mathrm{1}>\in\mathbb{N} \\ $$$${if}\:\:{y}=\mathrm{3}{x}\:\:{then}\:\:,\:{find}\:{the}\:{smallest}\: \\ $$$${value}\:{of}\:\:{x} \\ $$$${Q}\mathrm{2}\:\therefore\:{with}\:{the}\:{above}\:{conditions}\:,{what}\:{other}\:{values}\: \\ $$$${can}\:{be}\:{placed}\:\:{besides}\:{the}\:{number}\:``\:\mathrm{1}\:''\: \\ $$

Question Number 192087 Answers: 0 Comments: 3

Let {H_α } ∈ Ω, be a family of subgroup of a group G, then prove that ∩_(α ∈ Ω) H_α .

$$\mathrm{Let}\:\left\{\mathrm{H}_{\alpha} \right\}\:\in\:\Omega,\:\mathrm{be}\:\mathrm{a}\:\mathrm{family}\:\mathrm{of}\: \\ $$$$\mathrm{subgroup}\:\mathrm{of}\:\mathrm{a}\:\mathrm{group}\:\mathrm{G},\:\mathrm{then}\: \\ $$$$\mathrm{prove}\:\mathrm{that}\:\cap_{\alpha\:\in\:\Omega} \mathrm{H}_{\alpha} . \\ $$$$ \\ $$$$ \\ $$

Question Number 192062 Answers: 2 Comments: 3

prove it : times_n ; (√(4+(√(4+(√(4+...+(√4))))) )) < 3

$${prove}\:{it}\::\: \\ $$$$\:\:\:{times\_n}\:\:\:;\:\:\:\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+\sqrt{\mathrm{4}+...+\sqrt{\mathrm{4}}}}\:\:}\:<\:\mathrm{3} \\ $$

Question Number 192057 Answers: 1 Comments: 0

Simplify: (((√a) + (√a^2 ) + (√a^3 ) + (√a^4 ))/(((√a) + 1)∙(a + 1)))

$$\mathrm{Simplify}: \\ $$$$\frac{\sqrt{\mathrm{a}}\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{2}} }\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{3}} }\:\:+\:\:\sqrt{\mathrm{a}^{\mathrm{4}} }}{\left(\sqrt{\mathrm{a}}\:\:+\:\:\mathrm{1}\right)\centerdot\left(\mathrm{a}\:\:+\:\:\mathrm{1}\right)} \\ $$

Question Number 192039 Answers: 0 Comments: 0

let x,y,z be positive real number such that x^4 +y^4 +z^4 = 1 find the minimum value of (x^3 /(1−x^8 )) + (y^3 /(1−y^8 )) + (z^3 /(1−z^8 ))

$$\mathrm{let}\:{x},{y},{z}\:\mathrm{be}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{number}\:\mathrm{such}\:\mathrm{that} \\ $$$$\:{x}^{\mathrm{4}} +{y}^{\mathrm{4}} +{z}^{\mathrm{4}} \:=\:\mathrm{1}\:\mathrm{find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value} \\ $$$$\mathrm{of} \\ $$$$\:\:\:\:\frac{{x}^{\mathrm{3}} }{\mathrm{1}−{x}^{\mathrm{8}} }\:+\:\frac{{y}^{\mathrm{3}} }{\mathrm{1}−{y}^{\mathrm{8}} }\:+\:\frac{{z}^{\mathrm{3}} }{\mathrm{1}−{z}^{\mathrm{8}} } \\ $$

Question Number 192009 Answers: 2 Comments: 0

if a>1 , show ((Σ_(k=1) ^(a^2 −1) (√(a+(√k))))/(Σ_(k=1) ^(a^2 −1) (√(a−(√k))))) = (√2) + 1

$$\:\:\:\:\:{if}\:\:{a}>\mathrm{1}\:,\:{show} \\ $$$$\:\:\:\:\:\frac{\underset{{k}=\mathrm{1}} {\overset{{a}^{\mathrm{2}} −\mathrm{1}} {\sum}}\:\:\sqrt{{a}+\sqrt{{k}}}}{\underset{{k}=\mathrm{1}} {\overset{{a}^{\mathrm{2}} −\mathrm{1}} {\sum}}\:\:\sqrt{{a}−\sqrt{{k}}}}\:\:\:=\:\:\:\sqrt{\mathrm{2}}\:\:+\:\:\mathrm{1} \\ $$

Question Number 191961 Answers: 0 Comments: 0

Question Number 191950 Answers: 1 Comments: 0

(√a) = 1 + (1/a) find: a^2 −a−(√a) = ?

$$\sqrt{\mathrm{a}}\:=\:\mathrm{1}\:+\:\frac{\mathrm{1}}{\mathrm{a}}\:\:\:\mathrm{find}:\:\:\:\mathrm{a}^{\mathrm{2}} −\mathrm{a}−\sqrt{\mathrm{a}}\:=\:? \\ $$

Question Number 191946 Answers: 1 Comments: 0

((fof^(−1) (5)+fof^(−1) (15))/(fof^(−1) (5)))=?

$$\frac{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)+\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{15}\right)}{\mathrm{fof}^{−\mathrm{1}} \left(\mathrm{5}\right)}=? \\ $$

Question Number 191947 Answers: 2 Comments: 0

prove that (√(a+b(√(a−b(√(a+b(√(a−b(√(...)))))))))) = (((√(4a−3b^2 ))+b)/2)

$${prove}\:{that} \\ $$$$\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{{a}+{b}\sqrt{{a}−{b}\sqrt{...}}}}}\:\:=\:\:\frac{\sqrt{\mathrm{4}{a}−\mathrm{3}{b}^{\mathrm{2}} }+{b}}{\mathrm{2}} \\ $$

Question Number 191867 Answers: 1 Comments: 0

Prove that if u=f(x^3 +y^3 ),where f is arbitry function then x^2 (∂u/∂y) = y^2 (∂u/∂x)

$${Prove}\:{that}\:{if}\:\:\:{u}={f}\left({x}^{\mathrm{3}} +{y}^{\mathrm{3}} \right),{where}\:{f}\:\:{is}\:{arbitry} \\ $$$${function}\:{then}\:\:\:\:{x}^{\mathrm{2}} \:\frac{\partial{u}}{\partial{y}}\:=\:{y}^{\mathrm{2}} \frac{\partial{u}}{\partial{x}} \\ $$

Question Number 191839 Answers: 1 Comments: 0

2^a = 3^b = 36^c then prove that ab = 2c(a + b).

$$\mathrm{2}^{{a}} \:=\:\mathrm{3}^{{b}} \:=\:\mathrm{36}^{{c}} \:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$${ab}\:=\:\mathrm{2}{c}\left({a}\:+\:{b}\right). \\ $$

Question Number 191833 Answers: 3 Comments: 0

Question Number 191830 Answers: 0 Comments: 0

Question Number 191798 Answers: 2 Comments: 2

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