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a \(\dfrac{6}{15}+\dfrac{11}{15}=\dfrac{17}{15}\)
b \(\dfrac{22}{77}-\dfrac{14}{77}=\dfrac{8}{77}\)
c \(\dfrac{11}{13}\times\dfrac{26}{31}=11\times\dfrac{2}{31}=\dfrac{22}{31}\)
d \(\dfrac{1}{2}\times3\times\dfrac{2}{5}=\dfrac{3}{5}\)
\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)
\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)
\(a.\left(\frac{6}{11}+\frac{5}{11}\right).\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}b.\frac{3}{5}\cdot\frac{7}{9}+\frac{3}{5}\cdot\frac{2}{9}=\frac{3}{5}\cdot\left(\frac{7}{9}+\frac{2}{9}\right)=\frac{3}{5}\cdot1=\frac{3}{5}\)
a) $\frac{{14}}{{18}}:\frac{8}{9} = \frac{7}{9}:\frac{8}{9} = \frac{7}{9} \times \frac{9}{8} = \frac{{63}}{{72}} = \frac{7}{8}$
b) $\frac{9}{6}:\frac{3}{{10}} = \frac{3}{2}:\frac{3}{{10}} = \frac{3}{2} \times \frac{{10}}{3} = \frac{{30}}{6} = 5$
c) $\frac{4}{5}:\frac{{10}}{{15}} = \frac{4}{5}:\frac{2}{3} = \frac{4}{5} \times \frac{3}{2} = \frac{{12}}{{10}} = \frac{6}{5}$
d) $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$
a, \(\frac{5}{11}\times\frac{7}{25}+\frac{15}{11}\times\frac{1}{5}\)
\(=\frac{5}{11}\times\frac{7}{25}+\frac{5}{11}\times\frac{3}{5}\)
\(=\frac{5}{11}\times\left(\frac{7}{25}+\frac{3}{5}\right)\)
\(=\frac{5}{11}\times\frac{22}{25}\)
\(=\frac{2}{5}\)
b, \(\frac{3}{7}\times\frac{25}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\frac{75}{19}-\frac{1}{7}\times\frac{18}{19}\)
\(=\frac{1}{7}\times\left(\frac{75}{19}-\frac{18}{19}\right)\)
\(=\frac{1}{7}\times3\)
\(=\frac{3}{7}\)
\(\frac{8}{15}:\frac{2}{11}+\frac{7}{15}:\frac{2}{11}\)
\(=\left(\frac{8}{15}+\frac{7}{15}\right):\frac{2}{11}\)
\(=1:\frac{2}{11}\)
\(=\frac{11}{2}\)
\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)
\(\frac{12}{5}+y-\frac{8}{9}=1-0\)
\(\frac{12}{5}-y+\frac{8}{9}=1\)
\(\frac{12}{5}-y=1-\frac{8}{9}\)
\(\frac{12}{5}-y=\frac{1}{9}\)
\(y=\frac{12}{5}-\frac{1}{9}\)
\(y=\frac{108}{45}-\frac{5}{45}\)
\(y=\frac{103}{45}\)
a, \(\frac{6}{7}.\frac{16}{15}.\frac{7}{6}.\frac{21}{32}=\frac{6}{7}.\frac{7}{6}.\frac{16}{15}.\frac{21}{32}\)=\(1.\frac{16}{15}.\frac{21}{32}=\frac{7}{5.2}=\frac{7}{10}\)
Phần b T2
c,\(\frac{7}{4}.\frac{11}{21}+\frac{11}{21}.\frac{5}{4}=\frac{11}{21}.\left(\frac{7}{4}+\frac{5}{4}\right)\)=\(\frac{11}{21}.3=\frac{11}{7}\)
dấu . là nhân hay là phần ngăn cách ở hàng phần nghìn thế
DẤU CHẤM LÀ PHẦN NGĂN CÁCH