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Question Number 148739 Answers: 1 Comments: 2

Question Number 148689 Answers: 2 Comments: 0

Solve for x∈R ((x^2 −x+1)/(x^2 +x+1)) = (√((x^3 −1)/(x^3 +1))) .

$$\:\mathrm{Solve}\:\mathrm{for}\:\mathrm{x}\in\mathrm{R} \\ $$$$\:\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{x}+\mathrm{1}}{\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{1}}\:=\:\sqrt{\frac{\mathrm{x}^{\mathrm{3}} −\mathrm{1}}{\mathrm{x}^{\mathrm{3}} +\mathrm{1}}}\:. \\ $$

Question Number 148733 Answers: 0 Comments: 0

tg(α)+tg(β)=4 cos(α)+cos(β)=(1/5) tg(α+β)=?

$${tg}\left(\alpha\right)+{tg}\left(\beta\right)=\mathrm{4} \\ $$$${cos}\left(\alpha\right)+{cos}\left(\beta\right)=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$${tg}\left(\alpha+\beta\right)=? \\ $$

Question Number 148732 Answers: 3 Comments: 0

show that 8^n −3^n is divisible by 5 for all natural number.

$$\:\:\:\:\:\:{show}\:{that}\: \\ $$$$\mathrm{8}^{{n}} \:−\mathrm{3}^{{n}} \:{is}\:{divisible}\:{by}\:\mathrm{5}\:\:{for}\:{all}\:{natural} \\ $$$${number}. \\ $$

Question Number 148677 Answers: 1 Comments: 0

A series of natural numbers are grouped as 1+(2+3)+(4+5+6)+... such that the rth group contains r terms. Show that the sum of the numbers in the (2r−1)th group is r^4 −(r−1)^4 .

$$\:\:\mathrm{A}\:\mathrm{series}\:\mathrm{of}\:\mathrm{natural}\:\mathrm{numbers}\:\mathrm{are}\: \\ $$$$\:\:\mathrm{grouped}\:\mathrm{as}\:\mathrm{1}+\left(\mathrm{2}+\mathrm{3}\right)+\left(\mathrm{4}+\mathrm{5}+\mathrm{6}\right)+... \\ $$$$\:\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:{rth}\:\mathrm{group}\:\mathrm{contains}\:{r}\: \\ $$$$\:\:\mathrm{terms}.\:\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the} \\ $$$$\:\:\mathrm{numbers}\:\mathrm{in}\:\mathrm{the}\:\left(\mathrm{2}{r}−\mathrm{1}\right){th}\:\mathrm{group}\:\mathrm{is} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{r}}^{\mathrm{4}} −\left(\boldsymbol{{r}}−\mathrm{1}\right)^{\mathrm{4}} . \\ $$

Question Number 148676 Answers: 1 Comments: 0

Find local minimum of function H(x)=((34)/(3+((2x)/(x^2 +3x+1)))) .

$$\mathrm{Find}\:\mathrm{local}\:\mathrm{minimum}\:\mathrm{of}\:\mathrm{function} \\ $$$$\:\mathrm{H}\left(\mathrm{x}\right)=\frac{\mathrm{34}}{\mathrm{3}+\frac{\mathrm{2x}}{\mathrm{x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{1}}}\:. \\ $$

Question Number 148674 Answers: 1 Comments: 0

Find max & min value of (1) f(x)=sin^3 x (√(1+cos^2 x)) +cos^3 x(√(1+sin^2 x)) . (2) f(x)=sin x (√(1+cos^2 x)) +cos x (√(1+sin^2 x)) . x ∈ R

$$\:\mathrm{Find}\:\mathrm{max}\:\&\:\mathrm{min}\:\mathrm{value}\:\mathrm{of} \\ $$$$\:\left(\mathrm{1}\right)\:\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:^{\mathrm{3}} \mathrm{x}\:\sqrt{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:+\mathrm{cos}\:^{\mathrm{3}} \mathrm{x}\sqrt{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:. \\ $$$$\left(\mathrm{2}\right)\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{sin}\:\mathrm{x}\:\sqrt{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}}\:+\mathrm{cos}\:\mathrm{x}\:\sqrt{\mathrm{1}+\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}\:. \\ $$$$\:\mathrm{x}\:\in\:\mathbb{R}\: \\ $$

Question Number 148655 Answers: 3 Comments: 0

lim_(n→∞) ∫_( 0) ^( 1) ((nx)/(1 + n^2 x^4 )) dx = ?

$$\underset{\boldsymbol{{n}}\rightarrow\infty} {{lim}}\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{{nx}}{\mathrm{1}\:+\:{n}^{\mathrm{2}} {x}^{\mathrm{4}} }\:{dx}\:=\:? \\ $$

Question Number 148645 Answers: 1 Comments: 0

u_(n+3) =((u_(n+2) +u_(n+1) +u_n )/3) , ∀n∈IN find u_n

$${u}_{{n}+\mathrm{3}} =\frac{{u}_{{n}+\mathrm{2}} +{u}_{{n}+\mathrm{1}} +{u}_{{n}} }{\mathrm{3}}\:,\:\forall{n}\in{IN} \\ $$$${find}\:{u}_{{n}} \: \\ $$

Question Number 148639 Answers: 1 Comments: 0

find singular point of this following and whats the type of singular point ? (1)f(z)=(1/(lnz)) (2)f(z)=((1−cos(z+i))/(z(z^2 +1)^2 )) (3)f(z)=((sinz)/(z^2 +z)) (4)f(z)=((sin2z)/z^2 )

$${find}\:{singular}\:{point}\:{of}\:{this}\:{following}\:{and} \\ $$$${whats}\:{the}\:{type}\:{of}\:{singular}\:{point}\:? \\ $$$$ \\ $$$$\left(\mathrm{1}\right){f}\left({z}\right)=\frac{\mathrm{1}}{{lnz}} \\ $$$$ \\ $$$$\left(\mathrm{2}\right){f}\left({z}\right)=\frac{\mathrm{1}−{cos}\left({z}+{i}\right)}{{z}\left({z}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$ \\ $$$$\left(\mathrm{3}\right){f}\left({z}\right)=\frac{{sinz}}{{z}^{\mathrm{2}} +{z}} \\ $$$$ \\ $$$$\left(\mathrm{4}\right){f}\left({z}\right)=\frac{{sin}\mathrm{2}{z}}{{z}^{\mathrm{2}} } \\ $$

Question Number 148636 Answers: 1 Comments: 1

find tylor series of f(z)=logz about z_o =−1+i

$${find}\:{tylor}\:{series}\:{of}\:{f}\left({z}\right)={logz}\: \\ $$$${about}\:{z}_{{o}} =−\mathrm{1}+{i} \\ $$

Question Number 148635 Answers: 0 Comments: 0

show that {a_n }:=(1/(1!))−(1/(2!))+(1/(3!))−(1/(4!))...+(((−1)^(n+1) )/(n!)) is a cauchy sequence. my attempt: let ε>0 we have ∣a_m −a_n ∣=∣(((−1)^(n+2) )/((n+1)!))+(((−1)^(n+3) )/((n+2)!))+...+(((−1)^(m+1) )/(m!))∣ ≤∣(((−1)^(n+2) )/((n+1)!))∣+∣(((−1)^(n+3) )/((n+2)!))∣+...+∣(((−1)^(m+1) )/(m!))∣ =(1/((n+1)!))+(1/((n+2)!))+...+(1/(m!))<(1/(n!))≤(1/n)<ε. (1/n)→0 as n→∞, so no matter any ε>0 (1/n)<ε eventually. as long as n^∗ >(1/ε) ∣a_m −a_n ∣<ε ∀n≥n^∗ .

Question Number 148638 Answers: 0 Comments: 1

find laurent series f(z)=((cos(iz))/z^n ) ,∣z−i∣>2

$${find}\:{laurent}\:{series}\:{f}\left({z}\right)=\frac{{cos}\left({iz}\right)}{{z}^{{n}} }\:\:,\mid{z}−{i}\mid>\mathrm{2} \\ $$

Question Number 148633 Answers: 0 Comments: 0