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a) $2 \times \frac{3}{{14}} = \frac{2}{1} \times \frac{3}{{14}} = \frac{{2 \times 3}}{{1 \times 14}} = \frac{6}{{14}} = \frac{3}{7}$
b) $3 \times \frac{4}{9} = \frac{3}{1} \times \frac{4}{9} = \frac{{3 \times 4}}{{1 \times 9}} = \frac{{12}}{9} = \frac{4}{3}$
c) $\frac{7}{{18}} \times 6 = \frac{7}{{18}} \times \frac{6}{1} = \frac{{7 \times 6}}{{18 \times 1}} = \frac{{42}}{{18}} = \frac{7}{3}$
d) $\frac{{19}}{{12}} \times 0 = \frac{{19}}{{12}} \times \frac{0}{1} = \frac{{19 \times 0}}{{12 \times 1}} = 0$
\(\dfrac{18}{6}=18:6\)
\(\dfrac{50}{10}=50:10\)
\(\dfrac{15}{15}=15:15\)
\(\dfrac{12}{24}=12:24\)
a) $1 - \frac{1}{3} = \frac{3}{3} - \frac{1}{3} = \frac{{3 - 1}}{3} = \frac{2}{3}$
b) $1 - \frac{6}{9} = \frac{9}{9} - \frac{6}{9} = \frac{{9 - 6}}{9} = \frac{3}{9} = \frac{1}{3}$
c) $2 - \frac{2}{5} = \frac{{10}}{5} - \frac{2}{5} = \frac{{10 - 2}}{5} = \frac{8}{5}$
a) $\frac{1}{6}:\frac{3}{7} = \frac{1}{6} \times \frac{7}{3} = \frac{7}{{18}}$
b) $\frac{5}{{12}}:\frac{1}{4} = \frac{5}{{12}} \times \frac{4}{1} = \frac{{5 \times 4}}{{12 \times 1}} = \frac{{5 \times 4}}{{4 \times 3 \times 1}} = \frac{5}{3}$
c) $\frac{4}{{15}}:\frac{8}{3} = \frac{4}{{15}} \times \frac{3}{8} = \frac{{4 \times 3}}{{15 \times 8}} = \frac{{4 \times 3}}{{5 \times 3 \times 4 \times 2}} = \frac{1}{{10}}$
d) $\frac{{18}}{5}:\frac{9}{{10}} = \frac{{18}}{5} \times \frac{{10}}{9} = \frac{{18 \times 10}}{{5 \times 9}} = \frac{{9 \times 2 \times 5 \times 2}}{{5 \times 9}} = 4$
Mình sửa lại đề xíu.
a) \(\frac{75}{100}+\frac{18}{21}+\frac{19}{32}+\frac{1}{4}+\frac{3}{21}+\frac{13}{32}=\frac{3}{4}+\frac{1}{4}+\frac{18}{21}+\frac{3}{21}+\frac{19}{32}+\frac{13}{32}=1+1+1=3\)
b) \(4\frac{2}{5}+5\frac{6}{9}+2\frac{3}{4}+\frac{3}{5}+\frac{1}{3}+\frac{1}{4}=4+\frac{2}{5}+\frac{3}{5}+5+\frac{2}{3}+\frac{1}{3}+2+\frac{3}{4}+\frac{1}{4}\)
\(=4+1+5+1+2+1=14.\)
c) \(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{41\cdot43}=\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+...+\frac{43-41}{41\cdot43}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{41}-\frac{1}{43}=\frac{1}{3}-\frac{1}{43}=\frac{43-3}{3\cdot43}=\frac{40}{129}.\)
\(\frac{3}{4}=3:4;\frac{10}{3}=10:3;\frac{9}{5}=9:5;\frac{18}{2}=18:2;\frac{71}{100}=71:100\)