\(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-....-\frac{1}{95.100}\)
Bài 1: \(1-\frac{1}{5.10}-\frac{1}{10.15}-\frac{1}{15.20}-...-\frac{1}{95.100}\)
ta có B = 1- 1/5.10 - 1/10.15 -.......- 1/95 .100
=> 5B = 5 -( 5/5.10+5/10.15 +....+ 5/95.100
= > 5B = 5 - ( 1/5 -1/100 )
=> 5B= 481/100
=> B = 481/500
1-\(\frac{1}{5.10}\)- \(\frac{1}{10.15}\)- \(\frac{1}{15.20}\)-.......- \(\frac{1}{95.100}\)
C = 1/5.10 + 1/10.15 + 1/15.20 +...+ 1/95.100
C=1/5.10+1/10.15+...+1/95.100
= 5/5.10+5/10.15+...+5/95.100
= 1/5-1/10+1/10-1/15+...+1/95-1/100
= 1/5-1/100
= 19/100
\(C=5\times\left(1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..+\frac{1}{95}-\frac{1}{100}\right)\)
\(C=5\times\left(1-\frac{1}{100}\right)\)
\(C=5\times\frac{99}{100}\)
\(C=\frac{99}{20}\)
C = \(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{95.100}\)
= \(\frac{1}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\right)\)
= \(\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)\)
= \(\frac{1}{5}.\frac{19}{100}=\frac{19}{500}\)
\(1-\frac{2}{5.10}-\frac{2}{10.15}-.....-\frac{2}{95.100}\)
Tinh tong
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Tính
1-1/5.10-1/10.15-1/15.20...-1/95.100
\(1-\frac{1}{5\cdot10}-\frac{1}{10\cdot15}-\frac{1}{15\cdot20}-...-\frac{1}{95\cdot100}\)
\(=1-\left(\frac{1}{5\cdot10}+\frac{1}{10\cdot15}+...+\frac{1}{95\cdot100}\right)\)
\(=1-\frac{1}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=1-\frac{1}{5}\left(\frac{1}{5}-\frac{1}{100}\right)=1-\frac{19}{500}=\frac{481}{500}\)
tinh gia tri cua bieu thuc sau roi ghi ket qua vao o
A=\(\sqrt[5]{\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2005.2010}}\)
A = \(\frac{1}{5.10}\)+ \(\frac{1}{10.15}\)+ \(\frac{1}{15.20}\)+ ...... + \(\frac{1}{995.1000}\)
\(A=\frac{1}{5}x\left(\frac{1}{5}+\frac{1}{10}+\frac{1}{10}+\frac{1}{995.1000}\right)\)
\(A=\frac{1}{5}x\left(\frac{1}{5}-\frac{1}{1000}\right)\)
\(A=\frac{1}{5}x\frac{199}{1000}\)
\(A=\frac{199}{5000}\)
Nếu muốn thì thử lại :
\(=\frac{1}{5}\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{10}+..+\frac{1}{995}-\frac{1}{1000}\right)...\)
\(=\frac{1}{5}\left(1-\frac{1}{1000}\right)=\frac{1}{5}\cdot\frac{995}{1000}\)
tự tính nốt nha
\(A=\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{995.1000}\)
\(5A=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{5}-\frac{1}{20}+...+\frac{1}{995}-\frac{1}{1000}\)
\(5A=\frac{1}{5}-\frac{1}{1000}\)
\(5A=\frac{199}{1000}\)
\(A=\frac{199}{1000}:5\)
\(A=\frac{199}{5000}\)
Tính tổng 100 số hạng đầu tiên của dãy:
a) \(\frac{1}{5.10};\frac{1}{10.15};\frac{1}{15.20};...\)
b)\(\frac{1}{6};\frac{1}{66};\frac{1}{176};\frac{1}{336};...\)
Tính nhanh
A = \(\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{90.95}+\frac{1}{95.100}\)
AD em làm nhanh hộ mik. mik sẽ tick 2 nick . 1 nick tên trưởng xóm hình như ko có dấu . và nick này nữa
\(A=\frac{1}{1.5}+\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{95.100}\)
\(\Rightarrow\)\(5A=1+\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{95.100}\)
\(=1+\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{95}-\frac{1}{100}\)
\(=1+\frac{1}{5}-\frac{1}{100}=\frac{119}{100}\)
\(\Rightarrow\)\(A=\frac{119}{500}\)
A=1/1.5+1/5.10+....+1/95.100
=(5/1.5+5/5.10+...+5/95.100):5
=(1-1/5+1/5-1/10+...+1/95-1/100):5
=(1-1/100):5
=99/100:5
=99/500
A=1-1/5+1/5-1/10+...+1/95-1/100
A=1-1/100=99/100
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