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Question Number 197891 Answers: 2 Comments: 0

Soit I=∫^( 1) _( 0) (√(t(√(t(1−t)))))dt Comment calculer I

$$\mathrm{Soit}\:\mathrm{I}=\underset{\:\mathrm{0}} {\int}^{\:\mathrm{1}} \sqrt{\mathrm{t}\sqrt{\mathrm{t}\left(\mathrm{1}−\mathrm{t}\right)}}\mathrm{dt} \\ $$$$\mathrm{Comment}\:\mathrm{calculer}\:\mathrm{I} \\ $$

Question Number 197917 Answers: 1 Comments: 1

Question Number 197916 Answers: 2 Comments: 0

Question Number 197882 Answers: 1 Comments: 0

Find the minimum value of ((5t^2 −8t+5)/((2+(√3))t^2 −2t+2−(√3))) where 2−(√3)<t<2+(√3).

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of} \\ $$$$\frac{\mathrm{5}{t}^{\mathrm{2}} −\mathrm{8}{t}+\mathrm{5}}{\left(\mathrm{2}+\sqrt{\mathrm{3}}\right){t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{2}−\sqrt{\mathrm{3}}}\:\:\mathrm{where}\:\mathrm{2}−\sqrt{\mathrm{3}}<{t}<\mathrm{2}+\sqrt{\mathrm{3}}. \\ $$

Question Number 197881 Answers: 1 Comments: 1

Question Number 197880 Answers: 1 Comments: 0

find the sum of infinite series (1/2^1 )∙(1/3^2 ) + (1/2^2 )∙(1/3^4 )(1^2 +2^2 +3^2 ) + (1/2^3 )∙(1/3^6 )(1^2 +2^2 +3^2 +...+7^2 )+ (1/2^4 )∙(1/3^8 )(1^2 +2^2 +3^2 +...+15^2 )+........

$$\mathrm{find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{infinite}\:\mathrm{series} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{1}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{4}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} \right)\:+\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{6}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{7}^{\mathrm{2}} \right)+ \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{4}} }\centerdot\frac{\mathrm{1}}{\mathrm{3}^{\mathrm{8}} }\left(\mathrm{1}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} +...+\mathrm{15}^{\mathrm{2}} \right)+........ \\ $$

Question Number 197876 Answers: 2 Comments: 1

Question Number 197874 Answers: 1 Comments: 0

Question Number 197865 Answers: 1 Comments: 0

Question Number 197860 Answers: 3 Comments: 0

((2^(17) +2^(16) +2^(15) +…+1)/(2^8 +2^7 +2^6 +…+1)) = ?

$$\:\:\:\:\frac{\mathrm{2}^{\mathrm{17}} +\mathrm{2}^{\mathrm{16}} +\mathrm{2}^{\mathrm{15}} +\ldots+\mathrm{1}}{\mathrm{2}^{\mathrm{8}} +\mathrm{2}^{\mathrm{7}} +\mathrm{2}^{\mathrm{6}} +\ldots+\mathrm{1}}\:=\:?\: \\ $$

Question Number 197851 Answers: 0 Comments: 0

If z_1 = (√((α−β)/2)) and z_2 = (√((α+β)/2)) . Show that ∣z_1 −z_2 ∣^2 + ∣z_1 +z_2 ∣ = 2∣z∣^2 +∣z_2 ∣^2 and deduce that ∣α+(√(α^2 −β^2 ))∣ + ∣α−(√(α^2 −β^2 ))∣ = ∣α+β∣ + ∣α−β∣ Thank you in advance

$$\mathrm{If}\:\mathrm{z}_{\mathrm{1}} \:=\:\sqrt{\frac{\alpha−\beta}{\mathrm{2}}}\:\mathrm{and}\:\mathrm{z}_{\mathrm{2}} \:=\:\sqrt{\frac{\alpha+\beta}{\mathrm{2}}}\:.\:\mathrm{Show} \\ $$$$\mathrm{that}\:\mid\mathrm{z}_{\mathrm{1}} −\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:+\:\mid\mathrm{z}_{\mathrm{1}} +\mathrm{z}_{\mathrm{2}} \mid\:=\:\mathrm{2}\mid\mathrm{z}\mid^{\mathrm{2}} +\mid\mathrm{z}_{\mathrm{2}} \mid^{\mathrm{2}} \:\mathrm{and} \\ $$$$\mathrm{deduce}\:\mathrm{that}\: \\ $$$$\mid\alpha+\sqrt{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\mid\:+\:\mid\alpha−\sqrt{\alpha^{\mathrm{2}} −\beta^{\mathrm{2}} }\mid\:=\:\mid\alpha+\beta\mid\:+\:\mid\alpha−\beta\mid \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{in}\:\mathrm{advance} \\ $$

Question Number 197850 Answers: 1 Comments: 0

Solve the equation: (2sin x − 1)(2cos 2x + 2sin x +1) = 3(1−2cos 2x)

$${Solve}\:{the}\:{equation}: \\ $$$$\left(\mathrm{2}{sin}\:{x}\:−\:\mathrm{1}\right)\left(\mathrm{2}{cos}\:\mathrm{2}{x}\:+\:\mathrm{2}{sin}\:{x}\:+\mathrm{1}\right)\:=\:\mathrm{3}\left(\mathrm{1}−\mathrm{2}{cos}\:\mathrm{2}{x}\right) \\ $$

Question Number 197848 Answers: 4 Comments: 0