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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) $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$
\(a,\frac{1}{2}\times\frac{4}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\div\frac{6}{10}\)
\(=\frac{2}{5}\times\frac{10}{6}\)
\(=\frac{2}{3}\)
\(b,\frac{24}{35}\div\left(\frac{4}{5}\times\frac{8}{7}\right)\)
\(=\frac{24}{35}\div\frac{32}{35}\)
\(=\frac{24}{35}\times\frac{35}{32}\)
\(=\frac{3}{4}\)
a: Ta có:
\(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)=\dfrac{2}{5}+\dfrac{2}{5}=\dfrac{4}{5}\)
\(\dfrac{4}{5}=\dfrac{4}{5}\). Vậy \(\left(\dfrac{2}{5}+\dfrac{1}{5}\right)+\dfrac{1}{5}=\dfrac{2}{5}+\left(\dfrac{1}{5}+\dfrac{1}{5}\right)\)
Ta có:
\(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{7}{9}+\dfrac{1}{9}=\dfrac{8}{9}\)
\(\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)=\dfrac{2}{9}+\dfrac{6}{9}=\dfrac{8}{9}\)
\(\dfrac{8}{9}=\dfrac{8}{9}\). Vậy \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}=\dfrac{2}{9}+\left(\dfrac{5}{9}+\dfrac{1}{9}\right)\)
b: \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{3}{3}+\dfrac{4}{3}=\dfrac{7}{3}\)
\(\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)=\dfrac{1}{3}+\dfrac{6}{3}=\dfrac{7}{3}\)
\(\dfrac{7}{3}=\dfrac{7}{3}\). Vậy \(\left(\dfrac{1}{3}+\dfrac{2}{3}\right)+\dfrac{4}{3}=\dfrac{1}{3}+\left(\dfrac{2}{3}+\dfrac{4}{3}\right)\)
Đề của anh bị sai mới đúng chứ ạ? Anh Đạt ghi là \(\left(\dfrac{2}{9}+\dfrac{5}{9}\right)+\dfrac{1}{9}\) chứ có phải \(\dfrac{2}{5}\) đâu ạ?
a) \(\dfrac{2}{4}\times\dfrac{9}{5}=\dfrac{1}{2}\times\dfrac{9}{5}=\dfrac{1\times9}{2\times5}=\dfrac{9}{10}\)
b) \(\dfrac{13}{8}\times\dfrac{5}{15}=\dfrac{13}{8}\times\dfrac{1}{3}=\dfrac{13}{24}\)
c) \(\dfrac{3}{9}\times\dfrac{6}{12}=\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{1}{3\times2}=\dfrac{1}{6}\)
a: =18/20=9/10
b: \(=\dfrac{13\cdot5}{8\cdot15}=\dfrac{13}{8}\cdot\dfrac{1}{3}=\dfrac{13}{24}\)
c: \(=\dfrac{1}{3}\cdot\dfrac{1}{2}=\dfrac{1}{6}\)
1 \(A=\left(1+\frac{1}{2}\right)\times\left(1+\frac{1}{3}\right)\times\left(1+\frac{1}{4}\right)\times.........\times\left(1+\frac{1}{2016}\right)\times\left(1+\frac{1}{2017}\right)\)
\(A=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times......\times\frac{2016}{2017}\times\frac{2018}{2017}\)
\(A=\frac{2018}{2}=1009\)
\(B=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+.......+\frac{2}{43.45}\)
\(B=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-......+\frac{1}{43}-\frac{1}{45}\)
\(B=\frac{1}{3}-\frac{1}{45}\)
\(B=\frac{14}{45}\)
2 \(\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{2018}\times\frac{2017}{47}\)
\(=\frac{2017}{2018}\times\frac{23}{47}+\frac{24}{47}\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\times\left(\frac{23}{47}+\frac{24}{47}\right)\)
\(=\frac{2017}{2018}\times1\)
=\(\frac{2017}{2018}\)
bạn nào xem giải thế có đúng ko
a) \(\frac{2}{5}+\frac{9}{15}=\frac{2}{5}+\frac{3}{5}=\frac{5}{5}=1\)
b) \(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{7}{6}\)
c) \(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{12}{8}=\frac{3}{2}\)
d) \(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{19}{42}\)
a) \(\frac{2}{5}+\frac{9}{15}\)=\(\frac{2}{5}+\frac{3}{5}=\frac{2+3}{5}=1\)
b)\(\frac{15}{45}+\frac{25}{30}=\frac{1}{3}+\frac{5}{6}=\frac{2}{6}+\frac{5}{6}=\frac{2+5}{6}=\frac{7}{6}\)
c)\(\frac{28}{32}+\frac{45}{72}=\frac{7}{8}+\frac{5}{8}=\frac{7+5}{8}=\frac{12}{8}=\frac{3}{2}\)
d)\(\frac{8}{28}+\frac{5}{30}=\frac{2}{7}+\frac{1}{6}=\frac{12}{42}+\frac{7}{42}=\frac{12+7}{42}=\frac{19}{42}\)
\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)
\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)
\(\frac{12}{5}+y-\frac{8}{9}=1-0\)
\(\frac{12}{5}-y+\frac{8}{9}=1\)
\(\frac{12}{5}-y=1-\frac{8}{9}\)
\(\frac{12}{5}-y=\frac{1}{9}\)
\(y=\frac{12}{5}-\frac{1}{9}\)
\(y=\frac{108}{45}-\frac{5}{45}\)
\(y=\frac{103}{45}\)
a: \(\dfrac{3}{9}\times\dfrac{5}{4}=\dfrac{5}{4}\times\dfrac{1}{3}=\dfrac{5\times1}{4\times3}=\dfrac{5}{12}\)
b: \(\dfrac{10}{15}\times\dfrac{3}{5}=\dfrac{3}{5}\times\dfrac{2}{3}=\dfrac{3\times2}{5\times3}=\dfrac{2}{5}\)
c: \(\dfrac{5}{8}\times\dfrac{4}{12}=\dfrac{5}{8}\times\dfrac{1}{3}=\dfrac{5}{3\times8}=\dfrac{5}{24}\)
d: \(\dfrac{9}{27}\times\dfrac{3}{21}=\dfrac{1}{7}\times\dfrac{1}{3}=\dfrac{1\times1}{7\times3}=\dfrac{1}{21}\)
a) \(\dfrac{25}{60}+6\times\dfrac{1}{8}=\dfrac{25}{60}+\dfrac{3}{4}=\dfrac{25}{69}+\dfrac{45}{60}=\dfrac{25+45}{60}=\dfrac{70}{60}=\dfrac{7}{6}\)
b) \(\dfrac{15}{9}\times\left(6:\dfrac{30}{9}\right)=\dfrac{15}{9}\times\left(6\times\dfrac{30}{9}\right)=\dfrac{15}{9}\times20=\dfrac{100}{3}\)