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Câu 1:
B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)
= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)
= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
1,
Tỉ số giữa 10 quyển và 15 quyển:
10: 15 = 2/3
Nếu chia đều thì mỗi bạn nhận đc:
[15x 2 + 10x3] : [2+3] = 12 [quyển]
Vậy:....................
2,
1/2 + 1/3 + 1/4 + ... + 1/50 = [1 - 1/2] + [1-2/3] + ... + [1 - 49/50]
= 1 - 1/2 + 1 - 2/3 + ... + 1 - 49/50
= [1 + 1 + 1 +... + 1] - [1/2+2/3+3/4+...+49/50]
= 49 - [1/2+2/3+3/4+...+49/50]
Vậy 1/2 + 1/3 + 1/4 + ... + 1/50 không là số tự nhiên
3,
1/42 + 1/52 + ... +1/1002 < 1/3.4 + 1/4.5 + 1/5.6 + ... + 1/99.100
<=> 1/42 + 1/52 + ... +1/1002 < 1/3 - 1/4 + 1/4 - 1/5 + ... + 1/99 - 1/100
<=> 1/42 + 1/52 + ... +1/1002 < 1/3 - 1/100
<=> E < 1/3 - 1/100
=> E < 1/3
Mà 1/3 - 1/100 = 97/300 > 1/5
=> 1/5 < E < 1/3
4, A:
2013/1 + 2014/2+2015/3+...+4023/2011+4024/2012 - 2012
= ( 2013/1 - 1)+(2014/2 - 1) + ( 2015/3 - 1)+...+ (4023/2011 - 1) + ( 4024/2012 - 1)
= 2012(1+1/2+1/3+...+ 1/2011+1/2012)
Vậy \(A=\frac{\text{(1+1/2+1/3+...+ 1/2011+1/2012)}}{\text{2012(1+1/2+1/3+...+ 1/2011+1/2012)}}=\frac{1}{2012}\)
Câu B mik sẽ làm sau, bây giờ mik bận
Tỉ số giữa 10 quyển và 15 quyển:
10:15=2/3
Vậy nếu chia cho cả lớp thì mõi bạn nhận được:
(15x2+10x3):5=12 quyển
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2013.2014}=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2013}-\frac{1}{2014}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2013}+\frac{1}{2014}-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2014}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2013}+\frac{1}{2014}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1007}\right)\)
\(=\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\)
Lại có B = \(\frac{1}{1008.2014}+\frac{1}{1009.2013}+\frac{1}{1010.2012}+...+\frac{1}{2014.1008}\)
=> 3022B = \(\frac{3022}{1008.2014}+\frac{3022}{1009.2013}+\frac{3022}{1010.2012}+...+\frac{3022}{2014.1008}\)
\(=\frac{1}{1008}+\frac{1}{2014}+\frac{1}{1009}+\frac{1}{2013}+\frac{1}{1010}+\frac{1}{2012}+...+\frac{1}{2014}+\frac{1}{1008}\)
\(=2.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
=> \(B=\frac{1}{1511}.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
Khi đó \(\frac{A}{B}=\frac{\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{1511}.\left(\frac{1}{1008}+\frac{1}{1009}+\frac{1}{1010}+...+\frac{1}{2013}+\frac{1}{2014}\right)}=\frac{1}{\frac{1}{1511}}=1511\)
=> \(\frac{A}{B}=1511\)
=> A/B là 1 số nguyên (đpcm)