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\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.......+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{2011}}\)
\(\Rightarrow2A-A=2-\frac{1}{2^{2012}}\)
\(\Rightarrow A=2-\frac{1}{2^{2012}}\)
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) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
Ta có: A = \(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
2A = \(2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
2A = \(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
2A - A = \(\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
A = \(2-\frac{1}{2^{2012}}\)
Đặt \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
\(\Rightarrow2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
- Bn lấy 2A - A = A là ra nhé :))
1) Ta có : \(\frac{x-2}{4}=\frac{5+x}{3}\)
\(\Rightarrow\left(x-2\right).3=\left(5+x\right).4\)
\(\Rightarrow3x-6=20+4x\)
\(\Rightarrow3x=26+4x\)
\(\Rightarrow3x=26+x+3x\)
\(\Rightarrow0=26+x\)
\(\Rightarrow x=0-26\)
\(\Rightarrow x=-26\)
2) Ta có : \(A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\)
\(\Rightarrow\frac{1}{A}=1+2+2^2+...+2^{2012}\)
\(\Rightarrow\frac{2}{A}=2+2^2+2^3+...+2^{2013}\)
\(\Rightarrow\frac{2}{A}-\frac{1}{A}=\left(2+2^2+2^3+...+2^{2013}\right)-\left(1+2+2^2+...+2^{2012}\right)\)
\(\Rightarrow\frac{1}{A}=2^{2013}+1\)
\(\Rightarrow A=\frac{1}{2^{2013}+1}\)
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu
\(2012+\frac{2012}{1+2}+\frac{2012}{1+2+3}+.....+\frac{2012}{1+2+3+....+2011}\)
\(=\frac{2012}{\frac{1\left(1+1\right)}{2}}+\frac{2012}{\frac{2\left(2+1\right)}{2}}+\frac{2012}{\frac{3\left(3+1\right)}{2}}+.....+\frac{2012}{\frac{2011\left(2011+1\right)}{2}}\)
\(=\frac{4024}{1.2}+\frac{4024}{2.3}+\frac{4024}{3.4}+.....+\frac{4024}{2011.2012}\)
\(=4024\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2011}-\frac{1}{2012}\right)\)
\(=4024\left(1-\frac{1}{2012}\right)\)
\(=4024.\frac{2011}{2012}\)
\(=4022\)
MS=\(\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{2}{2011}\right)+\left(1+\frac{1}{2012}\right)\)
=\(2013\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2012}\right)\)
=> \(x.\frac{1}{2013}=1\)
=>x=2013
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thanks