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\(\dfrac{3\dfrac{1}{3}+\dfrac{1}{5}+2\dfrac{7}{15}}{\left(\dfrac{3}{10}+\dfrac{1}{4}+\dfrac{7}{20}\right).\dfrac{5}{6}}=\dfrac{\dfrac{10}{3}+\dfrac{1}{5}+\dfrac{37}{15}}{\left(\dfrac{6}{20}+\dfrac{5}{20}+\dfrac{7}{20}\right).\dfrac{5}{6}}\)
\(=\dfrac{\dfrac{50}{15}+\dfrac{3}{15}+\dfrac{37}{15}}{\dfrac{18}{20}.\dfrac{5}{6}}=\dfrac{6}{\dfrac{3}{4}}=\dfrac{6.4}{3}=8\)
Đổi: 1 và 1/2=3/2
Số bé là: 8:(3-2)x2=16
Số lớm là: 8:(3-2)x3=24
**** bạn
Ta có: 1/3 ; 1/15 ; 1/35;...
<=> 1/1.3 ; 1/3.5 ; 1/5.7
=> chữ số thứ 100 là: 1/199.201
Ta có: \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{199.201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)
a,\(\frac{2}{3}-\frac{5}{7}\times\frac{14}{25}\)
=\(\frac{2}{3}-\frac{5\times7\times2}{7\times5\times5}\)
=\(\frac{2}{3}-\frac{2}{5}\)
=\(\frac{10}{15}-\frac{6}{15}\)
=\(\frac{4}{15}\)
b,\(\frac{-2}{5}\times\frac{5}{8}+\frac{5}{8}\times\frac{3}{5}\)
=\(\frac{5}{8}\times\left(\frac{-2}{5}+\frac{3}{5}\right)\)
=\(\frac{5}{8}\times\frac{1}{5}\)
=\(\frac{1}{8}\)
a) \(\frac{x}{9}=\frac{8}{6}\Rightarrow x=\frac{8}{6}.9=\frac{4}{3}.9=12\)
b) \(\frac{12}{x}=\frac{6}{7}\Rightarrow x=12:\frac{6}{7}=12.\frac{7}{6}=14\)
c) \(5\frac{2}{3}:x=3\frac{2}{3}-2\frac{1}{2}\Rightarrow\frac{17}{3}:x=\left(3-2\right)+\left(\frac{2}{3}-\frac{1}{2}\right)\)
\(\Rightarrow\frac{17}{3}:x=1+\frac{1}{6}=\frac{7}{6}\Rightarrow x=\frac{17}{3}:\frac{7}{6}=\frac{17}{3}.\frac{6}{7}=\frac{34}{7}\)
\(\frac{4}{3}+\frac{9}{8}+...+\frac{9801}{9800}\)
\(=1+\frac{1}{2^2-1}+1+\frac{1}{3^2-1}+...+1+\frac{1}{99^2-1}\)
\(=98+\frac{1}{1.3}+\frac{1}{2.4}+\frac{1}{3.5}+\frac{1}{4.6}+...+\frac{1}{97.99}+\frac{1}{98.100}\)
\(=98+\frac{1}{2}\left(\frac{2}{1.3}+\frac{2}{2.4}+\frac{2}{3.5}+\frac{2}{4.6}+...+\frac{2}{97.99}+\frac{2}{98.100}\right)\)
\(=98+\frac{1}{2}\left(\frac{3-1}{1.3}+\frac{4-2}{2.4}+\frac{5-3}{3.5}+\frac{6-4}{4.6}+...+\frac{99-97}{97.99}+\frac{100-98}{98.100}\right)\)
\(=98+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{97}-\frac{1}{99}+\frac{1}{98}-\frac{1}{100}\right)\)
\(=98+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{99}-\frac{1}{100}\right)\)
\(=98+\frac{14651}{19800}\)