Tính (theo mẫu).
Mẫu: \(2+\dfrac{1}{6}=\dfrac{12}{6}+\dfrac{1}{6}=\dfrac{13}{6};1-\dfrac{1}{4}=\dfrac{4}{4}-\dfrac{1}{4}=\dfrac{3}{4}\) |
a) \(1+\dfrac{4}{9}\) b) \(5+\dfrac{1}{2}\) c) \(3-\dfrac{5}{6}\) d) \(\dfrac{31}{7}-2\)
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a: \(\dfrac{1}{8}+\dfrac{5}{8}=\dfrac{1+5}{8}=\dfrac{6}{8}=\dfrac{3}{4}\)
b: \(\dfrac{1}{15}+\dfrac{4}{15}=\dfrac{1+4}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{5}{9}+\dfrac{7}{9}=\dfrac{5+7}{9}=\dfrac{12}{9}=\dfrac{4}{3}\)
d: \(\dfrac{23}{100}+\dfrac{27}{100}=\dfrac{23+27}{100}=\dfrac{50}{100}=\dfrac{1}{2}\)
a) \(\dfrac{3}{4}-\dfrac{1}{8}=\dfrac{6}{8}-\dfrac{1}{8}=\dfrac{6-1}{8}=\dfrac{5}{8}\)
b) \(\dfrac{2}{6}-\dfrac{5}{18}=\dfrac{6}{18}-\dfrac{5}{18}=\dfrac{6-5}{18}=\dfrac{1}{18}\)
c) \(\dfrac{2}{5}-\dfrac{3}{20}=\dfrac{8}{20}-\dfrac{3}{20}=\dfrac{8-3}{20}=\dfrac{5}{20}=\dfrac{1}{4}\)
a: \(\dfrac{15}{8}-\dfrac{13}{8}=\dfrac{15-13}{8}=\dfrac{2}{8}=\dfrac{1}{4}\)
b: \(\dfrac{7}{15}-\dfrac{2}{15}=\dfrac{7-2}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)
c: \(\dfrac{11}{12}-\dfrac{2}{12}=\dfrac{11-2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
d: \(\dfrac{19}{7}-\dfrac{5}{7}=\dfrac{19-5}{7}=\dfrac{14}{7}=2\)
a: \(=\dfrac{37}{4}+\dfrac{117}{16}+\dfrac{1}{4}=\dfrac{19}{2}+\dfrac{117}{16}=\dfrac{269}{16}\)
b: \(=1+\left(\dfrac{9}{10}+\dfrac{8}{10}\right):\dfrac{19}{6}=1+\dfrac{17}{10}\cdot\dfrac{6}{19}=\dfrac{146}{95}\)
c: \(=\dfrac{1}{4}-\dfrac{6}{4}+\dfrac{6}{5}=\dfrac{-5}{4}+\dfrac{6}{5}=\dfrac{-1}{20}\)
a) Ta có: \(a\left(-\dfrac{3}{2}\right)+a\cdot\dfrac{1}{4}-a\cdot\dfrac{5}{6}\)
\(=a\left(-\dfrac{3}{2}+\dfrac{1}{4}-\dfrac{5}{6}\right)\)
\(=a\left(\dfrac{-18}{12}+\dfrac{3}{12}-\dfrac{10}{12}\right)\)
\(=a\cdot\dfrac{-25}{12}\)(1)
Thay \(a=\dfrac{3}{5}\) vào biểu thức (1), ta được:
\(\dfrac{3}{5}\cdot\dfrac{-25}{12}=\dfrac{-75}{60}=\dfrac{-5}{4}\)
1: \(\dfrac{1}{2}+\dfrac{9}{10}+\dfrac{5}{6}-\dfrac{11}{14}-\dfrac{1}{3}+\dfrac{-4}{35}\)
\(=\left(\dfrac{1}{2}+\dfrac{5}{6}-\dfrac{1}{3}\right)+\dfrac{9}{10}-\left(\dfrac{11}{14}+\dfrac{4}{35}\right)\)
\(=\dfrac{3+5-2}{6}+\dfrac{9}{10}-\dfrac{55+8}{70}\)
\(=1+\dfrac{9}{10}-\dfrac{9}{10}\)
=1
a) \(1+\dfrac{4}{9}=\dfrac{9}{9}+\dfrac{4}{9}=\dfrac{9+4}{9}=\dfrac{13}{9}\)
b) \(5+\dfrac{1}{2}=\dfrac{10}{2}+\dfrac{1}{2}=\dfrac{10+1}{2}=\dfrac{11}{2}\)
c) \(3-\dfrac{5}{6}=\dfrac{18}{6}-\dfrac{5}{6}=\dfrac{18-5}{6}=\dfrac{13}{6}\)
d) \(\dfrac{31}{7}-2=\dfrac{31}{7}-\dfrac{14}{7}=\dfrac{31-14}{7}=\dfrac{17}{7}\)