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Answers ( 459 )
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    10 = 2.5

    20 = 22.5

    40 = 23.5

    => GCD\(\left(10;20;40\right)=2.5=10\)

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    4 = 2^2

    8 = 2^3

    10 = 2.5

    => BCNN\(\left(4,8,10\right)=2^3\cdot5=40\)

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    A = 4.225.3 = 2700

    B = 20.m.5

    But A=B => 20.m.5 = 2700

    => 100m = 2700

    => m = 27

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    80 + 20 + 40 + 50 + 60 + 20

    = 80 + 20 + 40 + 60 + 50 + 20

    = 100 + 100 + 70

    = 270

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    \(A=2\left(\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{15\cdot16}\right)\)

    \(=2\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)

    \(=2\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=2\cdot\dfrac{3}{16}=\dfrac{3}{8}\)

    Ngân Hà auto ask, auto answer. 

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    Ngô Tấn Đạt's answer should write in English, if you write in Vietnamese, please go to http:/olm.vn to answer 

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    We have:\(x+1=\dfrac{b^2+c^2-a^2}{2bc}+1=\dfrac{b^2+2bc+c^2-a^2}{2bc}=\dfrac{\left(b+c\right)^2-a^2}{2bc}\)Inference:

    \(y\left(x+1\right)=\dfrac{a^2-\left(b-c\right)^2}{\left(b+c\right)^2-a^2}\cdot\dfrac{\left(b+c\right)^2-a^2}{2ab}=\dfrac{a^2-\left(b-c\right)^2}{2ab}\)

    => P = x + y + xy 

    \(=x+y\left(x+1\right)=\dfrac{b^2+c^2-a^2}{2bc}+\dfrac{a^2-\left(b-c\right)^2}{2bc}=\dfrac{b^2+c^2-a^2+a^2-\left(b-c\right)^2}{2bc}=\dfrac{b^2+c^2-a^2+a^2-b^2-2bc+c^2}{2bc}=\dfrac{2c^2-2bc}{2bc}=\dfrac{2c^2}{2bc}-1=\dfrac{c}{b}-1\)

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    22.23.24.25 = 22+3+4+5 = 214 = 16384

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    1 = 1 + 0

    2 = 1 + 1

    4 = 2 + 2

    7 = 4 + 3

    Next numbers are:

    7 + 4 = 11

    11 + 5 = 16

    16 + 6 = 22

    22 + 7 = 29

    29 + 8 = 37

    37 + 9 = 46

    46 + 10 = 56

    56 + 11 = 67

    67 + 12 = 79

    79 + 13 = 92

    And the last is 106 = 92 + 16

    So A or their sum are:

    1 + 2 + 4 + 7 + 11 + 16 + 22 + 29 + 37 + 46 + 56 + 67 + 79 + 92 + 106 = 508

    OK???

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    \(k\left(k+1\right)\left(k+2\right)-\left(k-1\right)k\left(k+1\right)\)

    \(=k\left(k+1\right)\left(k+2-k+1\right)\)

    \(=k\left(k+1\right)\cdot3\)

    \(=3k\left(k+1\right)\)

    Are you OK?

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    \(\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+\dfrac{1}{3}\)

    \(=\left(\dfrac{2}{4}+\dfrac{3}{4}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\)

    \(=\dfrac{5}{4}+1=\dfrac{5}{4}+\dfrac{4}{4}=\dfrac{9}{4}\)

    My answer is faster than Help you solve math's answer.
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    There are 31 days in the January

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    B = 20172016.20162017

    = \(\left(2017^4\right)^{504}.2016^{2017}\)

    \(=\left(...1\right)^{504}\cdot...6\)

    \(=...1\cdot...6=...6\)

    So the last digit of the number B is 6

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    A= n3 + 5n

    = n[n2 + 5]

    1. Prove that A divisible by 2

    - If n is a even => A divisible by 2

    - If n is a odd => n2 is a odd => n2 + 5 divisible 2 => A divisible by 2

    So A divisible by 2

    2. Prove that: A divisible by 3

    - If n divisible by 3 => A divisible by 3

    - If n not divisible by 3 => n2 \(\equiv1\left(mod3\right)\)

    => n2 + 5 divisible by 3

    => A divisible by 3

    So A divisible by 3

    But \(\left(2,3\right)=1\)

    => A divisible by 2.3 = 5

    Or A divisible by 6

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    We have:

    \(n^5-n=n\left(n^4-1\right)=n\left(n^2+1\right)\left(n^2-1\right)\)

    1. Prove that: A divisible by 3

    - If n divisible by 3 => A divisible by 3

    - If n\(⋮̸\)3 => \(n^2\equiv1\left(mod3\right)\) => n2 - 1 divisible by 3 => A divisible by 3

    So A divisible by 3 

    2. Prove that A divisible by 5

    - If n divisible by 5 => A divisible by 5

    - If n \(⋮̸\)5 => \(n^2\equiv1,4\left(mod5\right)\)

     + If n2 \(\equiv1\left(mod5\right)\Rightarrow n^2-1⋮5\) => A divisible by 5

      + If n2 \(\equiv4\left(mod5\right)\)=> \(n^2+1⋮5\) => A divisible by 5

    So A divisible by 5

    \(But\left(3,5\right)=1\)

    => A divisible by 3.5 = 15

    Or A divisible by 15

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    A = {0; 1; 2; ...; 49}

    \(\Leftrightarrow A=\left\{x\in N\text{|}x< 50\right\}\)

    or \(A=\left\{x\in N\text{|}x\le49\right\}\)

    B = {11; 12; ... ; 99}

    \(\Leftrightarrow B=\left\{x\in N\text{|}10< x< 100\right\}\)

    or \(B=\left\{x\in N\text{|}11\le x\le99\right\}\)

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    a, 46+17+54

    = \(\left(46+54\right)+17\)

    = 100+17

    = 117

    b,

    4.35.25

    = \(\left(4\cdot25\right)\cdot35\)

    = 100.35 = 3500

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    \(1\sqrt{4}+1\sqrt{4}+\sqrt[1]{5+\dfrac{5}{6}}\)

    \(=1\cdot2+1\cdot2+\sqrt{5+\dfrac{5}{6}}\)

    \(=2+2+\sqrt{\dfrac{210}{6}}\)

    \(=4+\sqrt{\dfrac{210}{6}}\)

    \(=\dfrac{24}{6}+\dfrac{\sqrt{210}}{6}=\dfrac{24+\sqrt{210}}{6}\)

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    The number of pen he left is:

    167 - 7 - 20 = 140 [pens]

    Answer: 140 pens

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    Call the number to find is a

    => a = 342k + 47

    a = 18.19k + 2.19 + 9

       = 19[18k + 2] + 9

    => a divide by 19 residual 9

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