Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Nhận xét thấy:
\(\dfrac{1}{1.2}\)= 1-\(\dfrac{1}{2}\); \(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3}\);...
Ta có
A= 1-\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
A= 1- \(\dfrac{1}{6}\)
A= \(\dfrac{5}{6}\)
Vậy A= \(\dfrac{5}{6}\)
CAU NAY RAT DE NHA BAN
A=\(\dfrac{1}{1}\)-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{6}\)
A=1-\(\dfrac{1}{6}\)
=>A=\(\dfrac{5}{6}\)
d, `3,15+2,4=5,55`
e, \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}\)
f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`
\(h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)
\(e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}\)
\(f3.6\times\left(1.25+8.75\right)=3.6\times10=36\)
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}\)
=\(1-\dfrac{1}{5}\)
=\(\dfrac{4}{5}\)
`@` `\text {Ans}`
`\downarrow`
`a)`
\(A=\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-...-\dfrac{1}{9}\)
`=`\(\dfrac{1}{3}-\dfrac{1}{9}\)
`=`\(\dfrac{2}{9}\)
Vậy, \(A=\dfrac{2}{9}\)
`b)`
\(B=\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{23\cdot24}+\dfrac{1}{24\cdot25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\left(\dfrac{1}{6}-\dfrac{1}{6}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{25}\)
`=`\(\dfrac{1}{5}-\dfrac{1}{25}=\dfrac{4}{25}\)
Vậy, \(B=\dfrac{4}{25}\)
`c)`
\(C=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{100}\)
`=`\(1-\dfrac{1}{100}=\dfrac{99}{100}\)
Vậy, \(C=\dfrac{99}{100}\)
Ta có:
\(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}=1-\dfrac{1}{10}=\dfrac{9}{10}\)
\(M=\dfrac{1}{2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}\)
\(M=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}\)
\(M=1-\dfrac{1}{7}\)
\(M=\dfrac{6}{7}\)
tham khảo
https://hoc24.vn/cau-hoi/123134145156167.5003535458609#:~:text=l%C3%BAc%2021%3A02-,1,14,-12.3%2B13.4%2B14.5
vào đi
\(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}=\dfrac{1}{3}\) .-.
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\)
\(=\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{3-1}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{19\cdot20}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}\)
\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)
\(A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+....+\dfrac{1}{19\cdot20}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{19}-\dfrac{1}{20}\)
\(A=1-\dfrac{1}{20}\)
\(A=\dfrac{19}{20}\)
\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{2}-\dfrac{1}{10}\)
\(=\dfrac{2}{5}\)
A= \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ \(\dfrac{1}{3.4}\)+ \(\dfrac{1}{4.5}\)+ \(\dfrac{1}{5.6}\)
= 1-\(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)- \(\dfrac{1}{3}\)+ \(\dfrac{1}{3}\)- \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)- \(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)- \(\dfrac{1}{6}\)
= 1 - \(\dfrac{1}{6}\)= \(\dfrac{5}{6}\)
mk chỉ bt làm câu 1 thôi ak
mong bn thông cảm