Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Giải:
Đặt:
\(A=\dfrac{6}{1.3.7}+\dfrac{6}{3.7.9}+\dfrac{6}{7.9.13}+\dfrac{6}{9.13.15}+\dfrac{6}{13.15.19}\)
\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{8}{1.3.7}+\dfrac{8}{3.7.9}+\dfrac{8}{7.9.13}+\dfrac{8}{9.13.15}+\dfrac{8}{13.15.19}\right)\)
\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{1.3}-\dfrac{1}{3.7}+\dfrac{1}{3.7}-\dfrac{1}{7.9}+\dfrac{1}{7.9}-\dfrac{1}{9.13}+\dfrac{1}{9.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.19}\right)\)
\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{1.3}-\dfrac{1}{15.19}\right)\)
\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{3}-\dfrac{1}{285}\right)\)
\(\Leftrightarrow A=\dfrac{6}{8}.\dfrac{94}{285}\)
\(\Leftrightarrow A=\dfrac{47}{190}\)
Vậy ...
6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
\(=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)\)
\(=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)
\(=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)\)
\(=\frac{6}{8}\cdot\frac{94}{285}\)
\(=\frac{47}{190}\)
Bạn ơi 6/8 o đâu ra vậy bạn có thể làm rõ ràng ra được không
6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)=86(1.3.78+3.7.98+...+13.15.198)
=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)=86(1.31−3.71+3.71−7.91+...+13.151−15.191)
=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)=86⋅(31−2851)
=\frac{6}{8}\cdot\frac{94}{285}=86⋅28594
=\frac{47}{190}=19047
a,A=\(\frac{1}{2}+\frac{1}{2.3}+\frac{1}{3.4}+.........+\frac{1}{23.24}\)
A=\(\frac{1}{2}+\frac{2}{1}-\frac{1}{3}+\frac{3}{1}-\frac{1}{4}+......\frac{23}{1}-\frac{1}{24}\)
A=\(\frac{1}{2}-\frac{1}{24}\)
A=\(\frac{11}{24}\)
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
Chỉ cần để các thừa số ra ngoài rồi nhân các số mà bằng khoảng cách của mẫu lên tử là giải được
=3(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+..+1/49-1/51)
=3x50/51=50/17
theo tôi là thế còn các bn
M = \(3\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{49\cdot51}\right)\)\(=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
M = \(3\left(1-\frac{1}{51}\right)=3\cdot\frac{50}{51}=\frac{50}{17}\)
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{4}{4}\left(\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=\frac{3}{3}\left(\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\right)\)
\(B=2\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}=\frac{1}{9}\)
Trả lời:
\(A=\frac{15}{90.94}+\frac{15}{94.98}+...+\frac{15}{146.150}\)
\(A=\frac{15}{4}.\left(\frac{4}{90.94}+\frac{4}{94.98}+...+\frac{4}{146.150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{94}+\frac{1}{94}-\frac{1}{98}+...+\frac{1}{146}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\left(\frac{1}{90}-\frac{1}{150}\right)\)
\(A=\frac{15}{4}.\frac{1}{225}\)
\(A=\frac{1}{60}\)
\(B=\frac{6}{15.18}+\frac{6}{18.21}+...+\frac{6}{87.90}\)
\(B=2.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(B=2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(B=2.\frac{1}{18}\)
\(B=\frac{1}{9}\)