Toán 6 Phân số

[tex]1-\frac{1}{2}+...+\frac{1}{125}-\frac{1}{126}-(\frac{2}{128}+...+\frac{2}{252})[/tex] Bài 1
A=[tex]\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}[/tex]
2A=[tex]\frac{2}{2}+\frac{2}{2^2}+...+\frac{2}{2^{10}}[/tex]
2A=[tex]1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}[/tex]
2A-A=([tex]1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}[/tex])-([tex]\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}[/tex])
A=1-[tex]\frac{1}{2^{10}}[/tex]
A=[tex]\frac{1023}{1024}[/tex]
B=[tex]1+\frac{1}{3}+...+\frac{1}{3^5}[/tex]
3B=[tex]3+1+...\frac{1}{3^4}[/tex]
3B-B=([tex]3+1+...\frac{1}{3^4}[/tex])-([tex]1+\frac{1}{3}+...+\frac{1}{3^5}[/tex])
2B=[tex]3-\frac{1}{3^5}[/tex]
B=[tex]\frac{364}{243}[/tex]
=>B>A
Bài 2
A=[tex](1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1009})-\frac{1}{1010}+...+\frac{1}{2019}[/tex]
A-B=[[tex](1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1009})-\frac{1}{1010}+...+\frac{1}{2019}[/tex]]-[tex](\frac{1}{1010}+...+\frac{1}{2019})[/tex]
=[tex]1-\frac{1}{2}+...+\frac{1}{1009}-(\frac{2}{1010}+\frac{2}{1011}+...+\frac{2}{2018})[/tex]
=[tex]1-\frac{1}{2}+...+\frac{1}{1009}-(\frac{1}{505}+\frac{1}{506}+...+\frac{1}{1009})[/tex]
=[tex]1-\frac{1}{2}+...+\frac{1}{503}-\frac{1}{504}-(\frac{2}{506}+\frac{2}{507}+...+\frac{2}{1008})[/tex]
=[tex]1-\frac{1}{2}+...+\frac{1}{503}-\frac{1}{504}-(\frac{1}{253}+\frac{1}{254}+...+\frac{1}{504})[/tex]
=[tex]1-\frac{1}{2}+...+\frac{1}{251}-\frac{1}{252}-(\frac{2}{254}+\frac{2}{255}+...+\frac{2}{504})[/tex]
=[tex]1-\frac{1}{2}+...+\frac{1}{251}-\frac{1}{252}[/tex][tex]-(\frac{1}{127}+...+\frac{1}{252})[/tex]

não mình dừng ở đây nhưng sau 1 quá trình dài nữa thì
A-B=0
=>(A-B-1)^2021=-1
Không biết sai chỗ nào không

Nguyễn Thị Quỳnh Lan said:

Bài 1: So sánh A và B biết:
[tex]A= \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ... + \frac{1}{2^{10}}[/tex]
[tex]B= 1 + \frac{1}{3} + \frac{1}{3^{2}} + \frac{1}{3^{3}} + \frac{1}{3^{4}} + \frac{1}{3^{5}}[/tex]
Bài 2: Cho
[tex]A = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{2017} - \frac{1}{2018} + \frac{1}{2019}[/tex]
[tex]B = \frac{1}{1010} + \frac{1}{1011} + ... + \frac{1}{2018} + \frac{1}{2019}[/tex]
Hãy tính [tex](A - b - 1)^{2021}[/tex]

Em cảm ơn
@Magic Boy @Cute Boy @hoàng việt nam

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Bài 2:
[TEX]A = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... + \frac{1}{2017} - \frac{1}{2018} + \frac{1}{2019}[/TEX]
[TEX]=> A = (1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2019})-(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018})[/TEX]
[TEX]=> A = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{2017} + \frac{1}{2018} + \frac{1}{2019})-2(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018})[/TEX]
[TEX]=> A = (1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{2017} + \frac{1}{2018} + \frac{1}{2019})-(1+ \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... +\frac{1}{1009}=\frac{1}{1010} + \frac{1}{1011} + ... + \frac{1}{2018} + \frac{1}{2019}=B [/TEX]
[TEX]=>A-B=0=>A-B-1=-1=>(A-B-1)^2021=-1[/TEX]