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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) \(\left(x-5\right).\left(10-9\frac{40}{41}\right):\left(1-\frac{81}{82}\right):\left(1-\frac{204}{205}\right)=2050\)
\(\Rightarrow\left(x-5\right).\frac{1}{41}.82.205=2050\)
\(\Rightarrow\left(x-5\right).2.205=2050\Leftrightarrow x-5=2050:410=5\Leftrightarrow x=10\)
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
\(\frac{11}{9}+\frac{35}{18}\)
\(=\frac{19}{6}\)
\(\frac{22}{21}:\frac{1}{2}+\frac{25}{12}:\frac{1}{2}\)
\(\frac{44}{21}+\frac{25}{6}\)
\(\frac{263}{42}\)
\(\frac{18}{11}+\left(\frac{7}{4}-\frac{3}{5}\right):\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}:\frac{1}{2}\)
\(\frac{18}{11}+\frac{23}{20}\times\frac{2}{1}\)
\(\frac{18}{11}+\frac{23}{10}\)
\(\frac{433}{110}\)
\(\frac{1}{5.8}\)\(+\frac{1}{8.11}+\frac{1}{11.14}+...+\frac{1}{x\left(x+3\right)}=\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=3.\frac{98}{1545}\)
\(\Leftrightarrow\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}\)
\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{103}\)
\(\Leftrightarrow x+3=103\)
\(\Leftrightarrow x\)\(=103-3\)
\(\Leftrightarrow x\)\(=100\)
Vậy x = 100
~~~~~~~Hok tốt~~~~~~~~
ta có \(\frac{1}{5.8}+\frac{1}{8.11}+...\frac{1}{x.\left(x+3\right)}\)\(=\frac{1}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{x.\left(x+3\right)}\right)\)\(=\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)\)
\(\Rightarrow\frac{1}{3}.\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{98}{1545}\)
\(\Rightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{98}{1545}:\frac{1}{3}=\frac{98}{515}\)
\(\Rightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{98}{515}=\frac{1}{103}\)
\(\Rightarrow x+3=103\)
\(\Rightarrow x=100\)
nhớ k nha
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{7}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}.\frac{6}{7}=\frac{1}{7}\)
\(\frac{6:\frac{3}{5}-1\frac{1}{6}\cdot\frac{6}{7}}{4\frac{1}{5}\cdot\frac{10}{11}+5\frac{2}{11}}=1\)
(1-1/2).(1-1/3).(1-1/4).(1-1/5)=b/100
=> 1/2.2/3.3/4.4/5=b/100
=> 1/5=b/100
=> b=100:5=20