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Question Number 160746 Answers: 1 Comments: 0

Question Number 160744 Answers: 1 Comments: 1

(x^2 +x−12)^3 +(x^2 +3x−18)^2 = 9(x^2 −9)^2 x=?

${(x^{2} + x - 12)}^{3} + {(x^{2} + 3 x - 18)}^{2} = 9 {(x^{2} - 9)}^{2}$ $x = ?$

Question Number 160739 Answers: 0 Comments: 3

nature de cette integrale quequ′en soit le reel α ∫_1 ^(+oo) t^α e^(−t) dt

$n a t u r e d e c e t t e i n t e g r a l e {q u e q u}^{'} e n s o i t l e r e e l α$ $\int_{1}^{+ o o} t^{α} e^{- t} d t$

Question Number 160747 Answers: 1 Comments: 1

Question Number 160734 Answers: 0 Comments: 1

# Advanced Calculus # Φ = ∫_0 ^( 1) (((ln^ ( (1/(1− x)) ))/x) )^( 3) dx =^? 3 ( ζ (2 ) + ζ (3 )) −−−− solution−−−− Φ =^(I.B.P) [ (( 1)/(2x^( 2) )) ln^( 3) ( 1−x)]_0 ^1 +(3/2) ∫_0 ^( 1) (( ln^( 2) (1− x ))/(x^( 2) (1 − x ))) dx = (1/2) lim_( ξ →1^(− ) ) ((ln^( 3) ( 1− ξ ))/ξ^( 2) ) +(3/2)[∫_0 ^( 1) (( ln^( 2) ( 1− x ))/x)dx = 2 ζ (3)] + (3/2)[∫_0 ^( 1) (( ln^( 2) ( 1−x))/x^( 2) ) dx = (π^( 2) /3) = 2ζ (2 )] +(3/2)∫_0 ^( 1) (( ln^( 2) (1− x))/(1−x)) dx} =(1/2) lim_( ξ →1^( −) ) {((ln^( 3) ( 1−ξ ))/ξ^( 2) ) −ln^( 3) (1− ξ ) } +(3/2) (2ζ (3 )) +(3/2) ( 2ζ (2 )) = 3( ζ (3 ) + 3ζ (2 ) ) ■ m.n

$You can't use 'macro parameter character #' in math mode$ $Φ = \int_{0}^{1} {(\frac{{l n}^{} (\frac{1}{1 - x})}{x})}^{3} d x \overset{?}{=} 3 (ζ (2) + ζ (3))$ $- - - - s o l u t i o n - - - -$ $Φ \overset{I . B . P}{=} {[\frac{1}{2 x^{2}} {l n}^{3} (1 - x)]}_{0}^{1} + \frac{3}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x^{2} (1 - x)} d x$ $= \frac{1}{2} {l i m}_{ξ \to 1^{-}} \frac{{l n}^{3} (1 - ξ)}{ξ^{2}}$ $+ \frac{3}{2} [\int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x} d x = 2 ζ (3)]$ $+ \frac{3}{2} [\int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x^{2}} d x = \frac{π^{2}}{3} = 2 ζ (2)]$ $+ \frac{3}{2} \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{1 - x} d x}$ $= \frac{1}{2} {l i m}_{ξ \to 1^{-}} {\frac{{l n}^{3} (1 - ξ)}{ξ^{2}} - {l n}^{3} (1 - ξ)}$ $+ \frac{3}{2} (2 ζ (3)) + \frac{3}{2} (2 ζ (2))$ $= 3 (ζ (3) + 3 ζ (2)) ◼ m . n$

Question Number 160733 Answers: 0 Comments: 0

Question Number 160755 Answers: 0 Comments: 0

Question Number 160753 Answers: 1 Comments: 1

Question Number 160752 Answers: 1 Comments: 0

solve (d^2 y/dx^2 )−y=x^2 sin3x

$s o l v e$ $\frac{d^{2} y}{{d x}^{2}} - y = x^{2} s i n 3 x$

Question Number 160719 Answers: 1 Comments: 2

Question Number 160807 Answers: 0 Comments: 0

−1≤a_0 ≤b_0 ≤c_0 ≤1 ∀n∈N a_(n+1) =∫_(−1) ^1 min(x,b_n ,c_n )dx b_(n+1) =∫_(−1) ^1 mil(x,a_n ,c_n )dx c_(n+1) =∫_(−1) ^1 max(x,b_n ,a_n )dx mil(a,b,c) est le terme median de (a,b,c) nature de (a_n ),(b_n ),(c_n )

$- 1 ⩽ a_{0} ⩽ b_{0} ⩽ c_{0} ⩽ 1$ $\forall n \in N$ $a_{n + 1} = \int_{- 1}^{1} m i n (x, b_{n}, c_{n}) d x$ $b_{n + 1} = \int_{- 1}^{1} m i l (x, a_{n}, c_{n}) d x$ $c_{n + 1} = \int_{- 1}^{1} m a x (x, b_{n}, a_{n}) d x$ $m i l (a, b, c) e s t l e t e r m e m e d i a n d e (a, b, c)$ $n a t u r e d e (a_{n}), (b_{n}), (c_{n})$

Question Number 160806 Answers: 1 Comments: 0

Question Number 160712 Answers: 0 Comments: 0

Solve for real numbers: (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) y-1 ; (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) z-1 (((x^2 +3n^2 )/(4n^2 )))^4 = (2/n) x-1 n ∈ (0 ; ∞) fixed

$Solve for real numbers :$ ${(\frac{x^{2} + {3 n}^{2}}{{4 n}^{2}})}^{4} = \frac{2}{n} y - 1; {(\frac{x^{2} + {3 n}^{2}}{{4 n}^{2}})}^{4} = \frac{2}{n} z - 1$ ${(\frac{x^{2} + {3 n}^{2}}{{4 n}^{2}})}^{4} = \frac{2}{n} x - 1$ $n \in (0; \infty) fixed$

Question Number 160711 Answers: 2 Comments: 0

Be p a prime number , arbitrary. Solve on positive integers (x;y;z) { ((xy + z^2 = 3p + 4)),((x + yz^2 = p + 4)) :}

$Be p a prime number, arbitrary .$ $Solve on positive integers (x; y; z)$ ${\begin{cases} xy + z^{2} = 3 p + 4 \\ x + {yz}^{2} = p + 4 \end{cases}$

Question Number 160707 Answers: 0 Comments: 0

Question Number 160706 Answers: 4 Comments: 0

Question Number 160701 Answers: 1 Comments: 0

Question Number 160695 Answers: 0 Comments: 0

Question Number 160694 Answers: 1 Comments: 0

Prove that 1 + (1/( (√2))) + (1/( (√3))) + …+ (1/( (√n))) < 2(√n)

$P r o v e t h a t$ $1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots + \frac{1}{\sqrt{n}} < 2 \sqrt{n}$

Question Number 160689 Answers: 1 Comments: 0

Question Number 160682 Answers: 0 Comments: 3

Question Number 160677 Answers: 1 Comments: 1

{ ((((x+y)/(xyz)) = −(1/4))),((((y+z)/(xyz)) = −(1/(24)))),((((x+z)/(xyz)) = (1/(24)))) :}

${\begin{cases} \frac{x + y}{xyz} = - \frac{1}{4} \\ \frac{y + z}{xyz} = - \frac{1}{24} \\ \frac{x + z}{xyz} = \frac{1}{24} \end{cases}$

Question Number 160672 Answers: 3 Comments: 1

Calculate lim_(x→0) ((tgx^m )/((sin x)^n )), lim_(x→0) ((xcos x−x)/((e^x −1)ln (1+3x^2 )))

$C a l c u l a t e$ $\lim_{x \to 0} \frac{{t g x}^{m}}{{(\sin x)}^{n}},$ $\lim_{x \to 0} \frac{x \cos x - x}{(e^{x} - 1) \ln (1 + 3 x^{2})}$

Question Number 160671 Answers: 0 Comments: 0

please ∫((lnx)/(x+lnx))dx calculate.

$p l e a s e \int \frac{l n x}{x + l n x} d x c a l c u l a t e .$

Question Number 160665 Answers: 0 Comments: 2

In how many ways can you divide 20 students into 5 groups with 4 students in each group such that any two students don′t meet each other in more than one group.

$I n h o w m a n y w a y s c a n y o u d i v i d e 20$ $s t u d e n t s i n t o 5 g r o u p s w i t h 4 s t u d e n t s$ $i n e a c h g r o u p s u c h t h a t a n y t w o$ $s t u d e n t s {d o n}^{'} t m e e t e a c h o t h e r i n$ $m o r e t h a n o n e g r o u p .$

Question Number 160661 Answers: 0 Comments: 0

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