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a.A= 1/2 + 1/4+ 1/8+ 1/16 + 1/32 + 1/64 + 1/128 + 1/256 + 1/512
A = 1 - 1/2 + 1/2- 1/4 + 1/4 - 1/8 + 1/8 - 1/16 + 1/16 - 1/32 + 1/32 - 1/64 + 1/64 - 1/128 + 1/128 - 1/256 - 1/256 - 1/512
A = 1 - 1/512
A = 511/512
b. 1/2 + 1/6 + 1/12 + … + 1/110
= 1/1.2 + 1/2.3 + 1/3.4 + … + 1/10.11. (dấu . thay dấu x).
= 1/1 – 1/2 + 1/2 – 1/3 + 1/3 – 1/4 +…+ 1/10 – 1/11
= 1/1 – 1/11
= 10/11
Chúc bạn học giỏi nha!
a ) Đặt \(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+....+\frac{1}{512}\)
\(\Rightarrow A=\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
Nhân 2 vào hai vế của biểu thức A , ta được :
\(2A=2.\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow2A=1+\frac{1}{2^1}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
Lấy biểu thức 2A - A , ta được :
\(2A-A=\left(1+\frac{1}{2^1}+\frac{1}{2^2}+....+\frac{1}{2^8}\right)-\left(\frac{1}{2^1}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
\(\Rightarrow A=1-\frac{1}{2^9}\Rightarrow A=\frac{512}{512}-\frac{1}{512}=\frac{511}{512}\)
Vậy \(A=\frac{511}{512}\)
b ) Đặt \(B=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+....+\frac{1}{90}+\frac{1}{110}\)
\(\Rightarrow B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(\Rightarrow B=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(B=1-\frac{1}{11}=\frac{11}{11}-\frac{1}{11}=\frac{10}{11}\)
Vậy \(B=\frac{10}{11}\)
ta có:
1/6+1/12+1/20+1/30+.........+1/90+1/110
= 1/2x3+1/3x4+1/4x5+1/5x6+....+1/9x10+1/10x11
= 1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+....+1/9-1/10+1/10-1/11
=1/2-1/11=11/22-2/22=9/22
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\frac{1}{2}-\frac{1}{11}=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{98.99}+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
Đặt tổng trên là A . Ta có:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\)
\(2A-A=\left(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{1024}\right)\)
\(A=1-\frac{1}{1024}\)
\(A=\frac{1023}{1024}\)
Bài làm
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+...+\frac{1}{90}+\frac{1}{110}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{9.10}+\frac{1}{10.11}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(A=\frac{1}{1}-\frac{1}{11}\)
\(A=\frac{10}{11}\)
1/2=1/1.2
1/6=1/2.3
1/12=1/3.4
1/20=1/4.5
1/30=1/5.6
1/42=1/6.7
ta có 1/1.2+1/2.3+1/3.4+1/4.5+1/5.6+1/6.7
= 1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7
=1/1-1/7
=6/7
misa
Đặt tên biểu thức là A ta có :
\(A=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{10}-\frac{1}{11}\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+.....+\left(\frac{1}{10}-\frac{1}{10}\right)\)
\(=\left(\frac{1}{2}-\frac{1}{11}\right)+0+......+0\)
\(=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\)
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+...+\frac{1}{10.11}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{10}-\frac{1}{11}\)
=\(\frac{1}{2}-\frac{1}{11}\)
=\(\frac{9}{22}\)
Ta có:
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
\(\Rightarrow2A=2\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^9}\right)=1+\frac{1}{2}+...+\frac{1}{2^8}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^9}\right)\)
\(\Rightarrow A=1-\frac{1}{2^9}=1-\frac{1}{512}=\frac{511}{512}\)
Vậy giá trị biểu thức là \(\frac{511}{512}\)
b) Ta có:
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{90}+\frac{1}{110}=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=1-\frac{1}{11}=\frac{10}{11}\)
Vậy giá trị biểu thức là \(\frac{10}{11}\)
minh ko hiểu gi cả?