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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
Ta có:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}\)
=\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
=\(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1006}\right)\)
=\(\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}\)
=>\(\left(\frac{A}{B}\right)^{2013}\)=(\(\frac{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}{\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}}^{ }\))2013=12013=1
\(\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2013}{1}+\frac{2014}{2}+\frac{2015}{3}+...+\frac{4024}{2012}-2012}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\left(\frac{2013}{1}-1\right)+\left(\frac{2014}{2}-1\right)+\left(\frac{2015}{3}-1\right)+...+\left(\frac{4024}{2012}-1\right)}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2012}{3}+...+\frac{2012}{2012}}\)
\(=\frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}}{2012.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}\right)}\)
\(=\frac{1}{2012}\)
Ủng hộ mk nha ^_-
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=\left(1+\frac{1}{3}+\frac{1}{5}+.....+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2012}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+..+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\left(1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{1006}\right)\)
\(S=\frac{1}{1007}+\frac{1}{1008}+.....+\frac{1}{2012}+\frac{1}{2013}=P\)
=>S-P=0
=>(S-P)2016=0
Câu 1:
B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)
= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)
= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)
= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)
= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)
\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)
các bn ơi
giúp mk đi mà
+.+