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\(\frac{33}{2}+\frac{33}{6}+\frac{33}{18}+\frac{33}{54}+\frac{33}{162}+\frac{33}{486}\)
\(=\frac{33.3+33.3+33.3+33.3+33.3}{486}\)
\(=\frac{99.5}{486}\)
\(=\frac{495}{486}\)
Gọi \(A=\frac{33}{2}+\frac{33}{6}+...+\frac{33}{486}\)
\(A=33.\left[\left(\frac{1}{1.2}+\frac{1}{2.3}\right)+\left(\frac{1}{3.6}+\frac{1}{6.9}\right)\left(\frac{1}{9.18}+\frac{1}{18.27}\right)\right]\)
\(A=33.\left[\frac{2}{3}+\frac{2}{9}+\frac{2}{27}\right]\)
\(A=66.\left[\frac{9}{27}+\frac{3}{27}+\frac{1}{27}\right]\)
\(A=66.\frac{13}{27}\)
\(A=\frac{286}{9}\)
sai hay đúng cx ko biết nha
\(S=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{97.100}\)
\(S=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{97.100}\right)\)
\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{97}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(S=\frac{1}{3}.\frac{49}{100}=\frac{49}{300}\)
Ta có: \(S=\frac{1}{2.5}+\frac{1}{5.8}+....+\frac{1}{97.100}.\)
\(\Rightarrow3S=\frac{3}{2.5}+\frac{3}{5.8}+....+\frac{3}{97.100}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{97}-\frac{1}{100}\)
\(\Rightarrow3S=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(\Rightarrow S=\frac{49}{100}:3=\frac{49}{300}\)
Vậy \(S=\frac{49}{300}\)
CHÚC BẠN HỌC TỐT
Ta có :
\(A=\frac{1}{8.11}+\frac{1}{11.14}+..+\frac{1}{605.608}\)
\(3A=\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{605.608}\)
\(3A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{605}-\frac{1}{608}\)
\(3A=\frac{1}{8}-\frac{1}{608}\)
\(A=\frac{75}{608}:3\)
\(A=\frac{25}{608}\)
Ủng hộ mk nha !!! ^_^
Ta có: \(A=\frac{1}{8.11}+\frac{1}{11.14}+.....+\frac{1}{605.608}\)
\(\Rightarrow3A=\frac{3}{8.11}+\frac{3}{11.14}+....+\frac{3}{605.608}\)
\(\Rightarrow3A=\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+....+\frac{1}{605}-\frac{1}{608}\)
\(\Rightarrow3A=\frac{1}{8}-\frac{1}{608}=\frac{75}{608}\)
\(\Rightarrow A=\frac{75}{608}:3=\frac{25}{608}\)
Vậy \(A=\frac{25}{608}\)
Ủng hộ cho mik nha bạn?
A = \(\frac{1}{3}\left(\frac{3}{8.11}+\frac{3}{11.14}+.......+\frac{3}{605.608}\right)\)
= \(\frac{1}{3}\left(\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{605}-\frac{1}{608}\right)\)
= \(\frac{1}{3}\left(\frac{1}{8}-\frac{1}{608}\right)=\frac{1}{3}.\frac{76-1}{608}=\frac{1}{3}.\frac{75}{608}=\frac{25}{608}\)
Ta có:
\(A=\left(x-\frac{1}{2}\right).\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\left(\frac{1}{1}-\frac{1}{10}\right)=\frac{1}{3}\)
\(\Leftrightarrow A=\left(x-\frac{1}{2}\right).\frac{9}{10}=\frac{1}{3}\Leftrightarrow x-\frac{1}{2}=\frac{1}{3}.\frac{10}{9}\Leftrightarrow x=\frac{47}{54}\)
\(B=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{96.101}=\frac{1}{10.x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{96.101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{96}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{5}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{1}{10}-\frac{1}{x}\Leftrightarrow B=\frac{1}{5}.\frac{100}{101}=\frac{1}{10}-\frac{1}{x}\)
\(\Leftrightarrow B=\frac{1}{x}=\frac{1}{10}-\frac{20}{101}=-\frac{99}{1010}\Leftrightarrow x=-\frac{1010}{99}\)
c) Sai đề nhé bạn vì không có kết quả nên không tìm được x.
d) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
giai câu a
a) ta có (\(\frac{2}{11.13}\)+\(\frac{2}{13.15}\)+.....+\(\frac{2}{19.21}\))*462 - x =19
(\(\frac{1}{11}\)-\(\frac{1}{13}\)+\(\frac{1}{13}\)-\(\frac{1}{15}\)+....+\(\frac{1}{19}\)-\(\frac{1}{21}\)) * 462 -x=19
(\(\frac{1}{11}\)-\(\frac{1}{21}\))*462-x=19
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(=1\left(\frac{1}{1}-\frac{1}{16}\right)\)
\(=1.\frac{15}{16}=\frac{15}{16}\)
Ta có: \(\frac{1}{10.11}+\frac{1}{11.12}+....+\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+....+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{10}-\frac{1}{50}\)
\(=\frac{2}{25}\)
=> 2/25