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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