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Nguyễn Nhật Minh
04/04/2017 at 06:29
Answers
3
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Prove that: 

 \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\)(n \(\in\)N*)

Fractioninequality

  • ...
    ¤« 19/02/2018 at 21:29

    34+536+7144+...+2n+1n2(n+1)2

    =1−122+122−132+132−142+...+1(n−1)2+1n2

    =1−1n2<1

  • ...
    Lê Quốc Trần Anh Coordinator 12/02/2018 at 09:27

    \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

    \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{\left(n-1\right)^2}+\dfrac{1}{n^2}\)

    \(=1-\dfrac{1}{n^2}< 1\)

  • ...
    FC Alan Walker 21/02/2018 at 17:38

    Ta có: \(\dfrac{2n+1}{n^2\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-n^2}{n^2\left(n+1\right)^2}=\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

    Do đó \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}\)

           \(=1-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+\dfrac{1}{3^2}-\dfrac{1}{4^2}+...+\dfrac{1}{n^2}-\dfrac{1}{\left(n+1\right)^2}\)

           \(=1-\dfrac{1}{\left(n+1\right)^2}< 1\)

    Vậy \(\dfrac{3}{4}+\dfrac{5}{36}+\dfrac{7}{144}+...+\dfrac{2n+1}{n^2\left(n+1\right)^2}< 1\forall n\in\)N*

                   


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Nobita - Kun
01/08/2018 at 05:56
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2
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A little while after Larry's BINGO, Jane got her own BINGO. Her face flushed with excitement, she yelled out "BINGO!" With a smirk, she turned to Larry and said:

a. "My BINGO has all prime numbers."

b. "The numbers in my BINGO add up to 138."

c. "My BINGO is a vertical line."

d. "My BINGO uses the FREE space."

One statement was false; the rest were true. Where was Jane's BINGO?

  • ...
    sssssssssssssss 01/08/2018 at 07:17

    c. "My BINGO is a vertical line."

    Nobita - Kun selected this answer.
  • ...
    Nobita - Kun 01/08/2018 at 07:08

    a. "My BINGO has all prime numbers."

    b. "The numbers in my BINGO add up to 138."

    c. "My BINGO is a vertical line."

    d. "My BINGO uses the FREE space."

    Help me this posts


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FA FIFA Club World Cup 2018
16/01/2018 at 21:16
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0
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For quadrilateral ABCD , E is the intersection of AB and CD , F is intersection of AD and BC , I and K are respectively the midpoint of BD and AC .
a , The points M belong to the inner domain of the quadrilateral and have the property

\(S_{MAB}+S_{MCD}=\dfrac{1}{2}S_{ABCD}\) on what road ?

b , Call N is the midpoint of EF . Prove that the points I , K , N are linear .


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FA FIFA Club World Cup 2018
16/01/2018 at 21:19
Answers
1
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Find the largest value and the smallest value of :

\(K=\dfrac{x+1}{x^2+3}\)

  • ...
    Vũ Mạnh Hùng 03/02/2018 at 18:49

    knjip


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HỦY DIỆT THE WORLD
16/01/2018 at 21:50
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3
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Solve the equation :

\(\dfrac{x^2-5x}{x-5}=5\)

  • ...
    FA FIFA Club World Cup 2018 16/01/2018 at 21:54

    We have : \(\dfrac{x^2-5x}{x-5}=5\)

    \(\dfrac{x\left(x-5\right)}{x-5}=5\)

    \(x=5\)

    So the experiment of the above equation is \(S=\left\{5\right\}\)

    HỦY DIỆT THE WORLD selected this answer.
  • ...
    Fc Alan Walker 22/01/2018 at 12:28

    We have : x2−5xx−5=5

    x(x−5)x−5=5

    x=5

    So the experiment of the above equation is S={5}

  • ...
    FC Alan Walker 21/02/2018 at 17:52

    Condition: \(x\ne5\)

    We have: \(\dfrac{x^2-5x}{x-5}=5\Leftrightarrow\dfrac{x\left(x-5\right)}{x-5}=5\Leftrightarrow x=5\)

    So the experiment of the above equation is \(S=\left\{\phi\right\}\)


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FA Thần Tốc Độ
16/01/2018 at 22:10
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0
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1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 +5 x 6 + ... + 99 x 100 = 0

Find x :


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Thao Dola
13/03/2017 at 13:30
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0
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Prove that for every pair of positive integers n and N, there are consecutive positive intefers k,k+1,...,k+N such that \(\varphi\left(k\right),\varphi\left(k+1\right),...,\varphi\left(k+N\right)\)are all divisible by n, where \(\varphi\left(n\right)\) is as defined in Euler's theorem.

Primitive in Arithmetic Progression


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Nguyen Thi Bich Huong
11/04/2017 at 15:09
Answers
1
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Convert fractions into fractions and then perform calculations:

a)\(2\dfrac{1}{5}+3\dfrac{4}{5}+9\dfrac{6}{5}\)

b)\(5\dfrac{7}{9}\cdot6\dfrac{2}{4}\cdot4\dfrac{5}{8}\)

  • ...
    Nguyen Thi Thuy Linh 11/04/2017 at 15:15

    a)\(2\dfrac{1}{5}+3\dfrac{4}{5}+9\dfrac{6}{5}=\dfrac{11}{5}+\dfrac{19}{5}+\dfrac{51}{5}=\dfrac{11+19+51}{5}=\dfrac{81}{5}\)

    b)\(5\dfrac{7}{9}\cdot6\dfrac{2}{4}\cdot4\dfrac{5}{8}=\dfrac{52}{9}\cdot\dfrac{26}{4}\cdot\dfrac{37}{8}=\dfrac{52\cdot26\cdot37}{9\cdot4\cdot8}=\dfrac{50024}{288}=173\dfrac{25}{36}\)

    Nguyen Thi Bich Huong selected this answer.

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Amano Ichigo
12/09/2017 at 20:38
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9
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Guess the animals. Write about them.

1 big and tall / brown and red / big feet and big ears / Australia / run and jump

2 big and long / in water / swim and walk / dangerous / eat fish

games

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    Nguyễn Thị Thanh Hiền 12/12/2017 at 09:09

    1 Kangaroo

    2 Shark

  • ...
    Pham Hoang Nam 05/02/2018 at 13:16

    1. Kangaroo

    2. Shark

  • ...
    anh Hào Trần 22/02/2018 at 12:22

    1 Kangaroo

    2 Shark


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Nguyen Tuan Anh
14/03/2017 at 14:06
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0
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Let p an odd prime number. For any positive integer k, show that there exists a positive integer m such that the rightmost k digits of \(m^2\), when expressed in the base p, are all 1's.

Base n Representatioons


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Trigonometric
sin cos tan cot sinh cosh tanh
Lim-log

Combined operations

 

α β γ δ θ σ ∂ ε ω φ ϕ π μ λ Ψ ξ η χ ζ ι κ ν ψ Ω ρ τ υ Γ Δ Λ Φ Π Σ Υ Ξ ϑ Θ ς ϰ
∞ ⊻ ⩞ ⋎ ⋏ ≀ ∪ ⊎ ⋓ ∩ ⋒ ⊔ ⊓ ⨿ ⊗ ⊙ ⊚ ⊛ ⊘ ⊝ ⊕ ⊖ ⊠ ◯ ⊥
⇔ ⇒ ⇐ → ← ↔ ↑ ↓
Operations
+ - ÷ × ≠ = ⊂ ⊃ ⊆ ⊇ ≈ ∈ ∉ ∃ ∄ ≤ ≥ ± ∓ ≠ ∅ ≃ ≅ ≡ ⋮ ⋮̸ ∀
(□) [□] {□} |□|

The type of system

m×n 1×2 1×3 1×4 1×5 1×6
2×1 2×2 2×3 2×4 2×5 2×6
3×1 3×2 3×3 3×4 3×5 3×6
4×1 4×2 4×3 4×4 4×5 4×6
5×1 5×2 5×3 5×4 5×5 5×6
6×1 6×2 6×3 6×4 6×5 6×6

Recipe:

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