Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta nhận thấy mẫu số của các phân số có qui luật 1x3; 2x4; 3x5; 4x6...... => mẫu số của phân số thứ 98 là 98x100
\(\Rightarrow A=\frac{4}{3}x\frac{9}{8}x\frac{16}{15}x\frac{25}{24}x\frac{36}{35}x...x\frac{9801}{9800}\)
\(A=\frac{2x2x3x3x4x4x5x5x6x6x...x99x99}{1x2x3x3x4x4x5x5x...x96x96x97x97x98x98x99x100}=\frac{2x99}{100}=\frac{99}{50}=1\frac{49}{50}\)
\(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right):\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)\)
=\(\frac{1}{3}+\frac{1}{5}\)
=\(\frac{5}{3}\)
\(S=\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{20.22}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{20.22}\)
\(2S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{20}-\frac{1}{22}\)
\(2S=1-\frac{1}{22}=\frac{21}{22}\)
\(S=\frac{21}{22}:2=\frac{21}{44}\)
a) \(\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{5}-\frac{1}{10}\)
\(=\frac{1}{10}\)
b) \(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{998.1000}\)
\(=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{998}-\frac{1}{1000}\)
\(=\frac{1}{10}-\frac{1}{1000}\)
\(=\frac{99}{1000}\)
c) \(\frac{4}{1.2}+\frac{4}{2.3}+\frac{4}{3.4}+...+\frac{4}{69.90}\)
\(=4.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{89.90}\right)\)
\(=4.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{89}-\frac{1}{90}\right)\)
\(=4.\left(1-\frac{1}{90}\right)\)
\(=4.\frac{89}{90}\)
\(=\frac{178}{45}\)
_Chúc bạn học tốt_
\(=\frac{6}{5}\times\frac{7}{6}\times...\times\frac{11}{10}\)(lại lỗi đề)
\(=\frac{6×7×...×11}{5×6×...×10}\)
\(=\frac{11}{5}\)
\(1\frac{1}{5}\cdot1\frac{1}{6}\cdot1\frac{1}{7}\cdot1\frac{1}{8}\cdot1\frac{1}{9}\cdot1\frac{1}{10}\)
\(=\frac{6}{5}\cdot\frac{7}{6}\cdot\frac{8}{7}\cdot\frac{9}{8}\cdot\frac{10}{9}\cdot\frac{11}{10}\)
\(=\frac{6\cdot7\cdot8\cdot9\cdot10\cdot11}{5\cdot6\cdot7\cdot8\cdot9\cdot10}\)
\(=\frac{11}{5}\)
\(=\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+...+\frac{4}{18x20}\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2x\left(\frac{1}{2}-\frac{1}{20}\right)\\ =2x\frac{9}{20}\\ =\frac{9}{10}\)
\(S=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+...+\frac{1}{224}\)
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{14.16}\)
\(2S=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{14.16}\)
\(2S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{16}\)
\(2S=\frac{1}{2}-\frac{1}{16}\)
\(S=\frac{7}{16}:2=\frac{7}{32}\)
Ủng hộ mk nha !!! ^_^
\(S=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{14.16}\)
\(S=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{14}-\frac{1}{16}\)
\(S=\frac{1}{2}-\frac{1}{16}\)
\(S=\frac{7}{16}\)
\(S=\frac{1}{8}+\frac{1}{24}+...+\frac{1}{224}\)
\(\Rightarrow\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{14.16}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{14.16}\right)\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{14}-\frac{1}{16}\right)\)
\(\Rightarrow\frac{1}{2}\left(1-\frac{1}{16}\right)\)
\(\Rightarrow\frac{1}{2}.\frac{15}{16}\)\(\Rightarrow S=\frac{15}{32}\)
A đù làm nhầm =)) :v
\(S=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+....+\frac{1}{224}\)
\(=\frac{1}{2.4}+\frac{1}{4.6}+....+\frac{1}{14.16}\)
\(=\frac{1}{2}\left(\frac{2}{2.4}+....+\frac{2}{14.16}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{4}+\frac{1}{4}-......+\frac{1}{14}-\frac{1}{16}\right)\)
\(=\frac{1}{2}\left(1-\frac{1}{16}\right)\)
\(=\frac{7}{32}\)
7/32 nhe
\(S=\frac{1}{8}+\frac{1}{24}+\frac{1}{48}+....+\frac{1}{224}\)
\(S=\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+.....+\frac{1}{14\cdot16}\)
\(2S=\frac{2}{2\cdot4}+\frac{2}{4\cdot6}+\frac{2}{6\cdot8}+.....+\frac{2}{14\cdot16}\)
\(2S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+.....+\frac{1}{14}-\frac{1}{16}\)
\(2S=1-\frac{1}{16}\)
\(S=\frac{15}{16}:2\)
\(S=\frac{15}{32}\)