Tính tổng :
A = \(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
B = \(\frac{12}{84}+\frac{12}{210}+\frac{12}{390}+\frac{12}{2100}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=2\left(\frac{1}{30}+\frac{1}{42}+.....+\frac{1}{380}\right)=2\left(\frac{6-5}{5.6}+\frac{7-6}{6.7}+.....+\frac{20-19}{20.19}\right)=2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)=2\left(\frac{1}{5}-\frac{1}{20}\right)=\frac{3}{10}\)
\(B=\frac{12}{84}+\frac{12}{210}+.....+\frac{12}{2100}=\frac{4}{28}+\frac{4}{70}+.....+\frac{4}{700}=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{1}{25.28}\right)=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-.....-\frac{1}{28}\right)=\frac{4}{2}.\frac{6}{28}=\frac{3}{7}\)
A=1/15+1/21+1/28+....+1/190
1/2A=1/30+1/42+1/56+.....+1/380
1/2A=1/5.6+1/6.7+1/7.8+....+1/19.20
1/2A=1/5-1/6+1/6-1/7+1/7-1/8+......+1/19-1/20
1/2A=1/5-1/20
1/2A=3/20
A=3/20:1/2
A=3/10
\(F=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(\Rightarrow\)\(\frac{1}{2}F=\frac{1}{2}.\left(\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\right)\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+...+\frac{1}{380}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{1}{5}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{4}{20}-\frac{1}{20}\)
\(\Rightarrow\) \(\frac{1}{2}F=\frac{3}{20}\)
\(\Rightarrow\)\(F=\frac{3}{20}\div\frac{1}{2}\)
\(\Rightarrow\) \(F=\frac{3}{20}.2\)
\(\Rightarrow\)\(F=\frac{3}{10}\)
\(F=\frac{1}{15}+\frac{ 1}{21}+...+\frac{1}{190}\)
\(F=\frac{2}{30}+\frac{2}{21}+...+\frac{2}{380}\)
\(F=\frac{2}{5.6}+...+\frac{2}{19.20}\)
\(F=2.\left(\frac{1}{5.6}+...+\frac{1}{19.20}\right)\)
\(F=2.\left(\frac{1}{5}-\frac{1}{6}+...+\frac{1}{19}-\frac{1}{20}\right)\)
\(F=2\left[\frac{1}{5}-\left(\frac{1}{6}-\frac{1}{6}\right)-...-\left(\frac{1}{19}-\frac{1}{19}\right)-\frac{1}{20}\right]\)
\(F=2.\left(\frac{1}{5}-\frac{1}{20}\right)\)
\(F=2.\frac{3}{20}\)
\(F=\frac{6}{20}=\frac{3}{10}\)
\(G=\frac{12}{84}+\frac{12}{210}+...+\frac{12}{2100}\)
\(G=\frac{4}{28}+\frac{4}{70}+...+\frac{4}{700}\)
\(G=\frac{4}{4.7}+\frac{4}{7.10}+...+\frac{4}{25.28}\)
\(G=\frac{4}{3}.\left(\frac{3}{4.7}+...+\frac{3}{25.28}\right)\)
\(G=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{7}+...+\frac{1}{25}-\frac{1}{28}\right)\)
\(G=\frac{4}{3}.\left(\frac{1}{4}-\frac{1}{28}\right)\)
\(G=\frac{4}{3}.\frac{6}{28}\)
\(G=\frac{2}{7}\)
Tổng của G và F là : \(\frac{3}{10}+\frac{2}{7}=\frac{21}{70}+\frac{20}{70}=\frac{41}{70}\)
Rút gọn biểu thức cho 3 ta được biểu thức mới là
\(\frac{4}{28}+\frac{4}{70}+\frac{4}{130}+...\frac{4}{700}\)
= \(\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{25.28}\)
=\(\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
= \(\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{25}-\frac{1}{28}\right)\)
= \(\frac{4}{3}\left(\frac{1}{4}-\frac{1}{28}\right)\)
= \(\frac{4}{3}.\frac{3}{14}\)
= \(\frac{2}{7}\)
bạn tách mẫu ra rồi lấy 12*với các tổng
rồi khử liên tiếp đc kết quả thì nhân với 12
\(\left(\frac{-1}{4}+\frac{7}{33}-\frac{5}{3}\right)-\left(\frac{-5}{4}+\frac{6}{11}-\frac{48}{49}\right)=\left(\frac{-1}{4}-\frac{16}{11}\right)-\left(-\frac{31}{44}-\frac{48}{49}\right)=-\frac{1}{4}-\frac{16}{11}+\frac{31}{44}+\frac{48}{49}=-\frac{1}{49}\)
\(A=\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+...+\frac{1}{190}\)
\(A=\frac{2}{30}+\frac{2}{42}+\frac{2}{56}+...+\frac{2}{380}\) ( nhân cả tử và mẫu với 2 )
\(A=\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{19.20}=2\left(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{19.20}\right)\)
A = \(2\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{20}\right)=2\left(\frac{1}{5}-\frac{1}{20}\right)=2.\frac{3}{20}=\frac{3}{10}\)
B = \(\frac{12}{84}+\frac{12}{210}+\frac{12}{390}+...+\frac{12}{2100}\)
\(B=\frac{4}{28}+\frac{4}{70}+\frac{4}{130}+...+\frac{4}{700}\) ( chia cả tử và mẫu của mỗi phân số cho 3 )
B = \(\frac{4}{4.7}+\frac{4}{7.10}+\frac{4}{10.13}+...+\frac{4}{25.28}=\frac{4}{3}\left(\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{25.28}\right)\)
B = \(\frac{4}{3}\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{25}-\frac{1}{28}\right)=\frac{4}{3}\left(\frac{1}{4}-\frac{1}{28}\right)=\frac{4}{3}.\frac{6}{28}=\frac{2}{3}\)
B = \(\frac{12}{84}+\frac{12}{210}+\frac{12}{390}+...+\frac{12}{2100}\)
Mik sửa lại đề bài