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
\(S=\frac{6}{2.5}+\frac{6}{5.8}+.......+\frac{6}{29.32}\)
\(S=2\left(\frac{3}{2.5}+\frac{3}{5.8}+......+\frac{3}{29.32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+......+\frac{1}{29}-\frac{1}{32}\right)\)
\(S=2\left(\frac{1}{2}-\frac{1}{32}\right)\)
\(S=2.\frac{15}{32}\)
\(S=\frac{15}{16}< 1\RightarrowĐPCM\)
Vậy \(S=\frac{15}{16}\)
a) Ý bạn là: \(S_1=\frac{3}{4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{40\cdot43}\)đúng không?
\(S_1=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}\)
\(S_1=1-\frac{1}{43}< 1\left(đpcm\right)\)
b) \(S_2=\frac{6}{2\cdot5}+\frac{6}{5.8}+\frac{6}{8\cdot11}+...+\frac{6}{29\cdot32}\)
=>\(\frac{S_2}{2}=\frac{3}{2\cdot5}+\frac{3}{5.8}+\frac{3}{8\cdot11}+...+\frac{3}{29\cdot32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{29}-\frac{1}{32}\)
\(\frac{S_2}{2}=\frac{1}{2}-\frac{1}{32}=\frac{16}{32}-\frac{1}{32}=\frac{15}{32}\)
=>\(S_2=\frac{15}{32}\cdot2=\frac{15}{16}< 1\left(đpcm\right)\)
\(\frac{\frac{6}{13}-\frac{6}{23}+\frac{6}{33}-\frac{6}{43}}{\frac{5}{13}-\frac{5}{23}+\frac{5}{33}-\frac{5}{43}}\)
= \(\frac{6.\left(\frac{1}{13}-\frac{1}{23}+\frac{1}{33}-\frac{1}{43}\right)}{5.\left(\frac{1}{13}-\frac{1}{23}+\frac{1}{33}-\frac{1}{43}\right)}\)
= \(\frac{6}{5}\)
k cho mình nhé
