Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
\(\frac{11}{9}+\frac{35}{18}\)
\(=\frac{19}{6}\)
\(\frac{22}{21}:\frac{1}{2}+\frac{25}{12}:\frac{1}{2}\)
\(\frac{44}{21}+\frac{25}{6}\)
\(\frac{263}{42}\)
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)
b)\(\left(2016.1017+2017.2018\right).\left(1+\frac{1}{2}:\frac{3}{2}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right)\left(1+\frac{1}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(\left(2016.2017+2017.2018\right).0\)
\(=0\)
a. \(\frac{-3}{2}-2x+\frac{3}{4}=-22\)2
=> \(-2x=-22+\frac{3}{2}-\frac{3}{4}\)
=> \(-2x=\frac{-85}{4}\)
=> \(x=\frac{-85}{4}:\left(-2\right)\)
=> \(x=\frac{85}{8}\)
b. \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\left(\frac{3}{-2}-\frac{10}{3}\right)=\frac{2}{5}\)
=> \(\left(\frac{-2}{3}x-\frac{3}{5}\right).\frac{-29}{6}=\frac{2}{5}\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{2}{5}:\left(\frac{-29}{6}\right)\)
=> \(\frac{-2}{3}x-\frac{3}{5}=\frac{-12}{145}\)
=> \(\frac{-2}{3}x=\frac{-12}{145}+\frac{3}{5}\)
=> \(\frac{-2}{3}x=\frac{15}{29}\)
=> x = \(\frac{15}{29}:\frac{-2}{3}\)
=> x = \(\frac{-45}{58}\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)