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)  $\frac{1}{6}:\frac{{21}}{9} = \frac{1}{6}:\frac{7}{3} = \frac{1}{6} \times \frac{3}{7} = \frac{3}{{42}} = \frac{1}{{14}}$

\(1-\left(\frac{12}{5}+y=\frac{8}{9}\right):\frac{16}{9}=0\)

\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\times\frac{16}{9}\)

\(1-\left(\frac{12}{5}+y-\frac{8}{9}\right)=0\)

\(\frac{12}{5}+y-\frac{8}{9}=1-0\)

\(\frac{12}{5}-y+\frac{8}{9}=1\)

\(\frac{12}{5}-y=1-\frac{8}{9}\)

\(\frac{12}{5}-y=\frac{1}{9}\)

\(y=\frac{12}{5}-\frac{1}{9}\)

\(y=\frac{108}{45}-\frac{5}{45}\)

\(y=\frac{103}{45}\)

a) \(\left(\frac{5}{12}+\frac{1}{4}\right)\cdot\frac{6}{4}=\left(\frac{5}{12}+\frac{3}{12}\right)\cdot\frac{3}{2}=\frac{8}{12}\cdot\frac{3}{2}\)\(=\frac{24}{24}=1\)

b) \(\left(\frac{5}{3}-\frac{7}{6}\right):\frac{2}{5}=\left(\frac{10}{6}-\frac{7}{6}\right):\frac{2}{5}=\frac{3}{6}:\frac{2}{5}\)\(=\frac{1}{2}:\frac{2}{5}=\frac{5}{4}\)

a.(5/12+1/4)+6/4

=(5/12+3/12)+6/4

=8/12+6/4

=8/12+18/12

=20/12

=5/3 

b.(5/3-7/6):2/5

=(30/18-21/18):2/5

=9/18:2/5

=1/2:2/5

=1/2*5/2

=5/4

Ta có: \(\frac56=1-\frac16;\frac78=1-\frac18;\frac{9}{10}=1-\frac{1}{10}\)

Ta có: 6<8<10

=>\(\frac16>\frac18>\frac{1}{10}\)

=>\(-\frac16<-\frac18<-\frac{1}{10}\)

=>\(-\frac16+1<-\frac18+1<-\frac{1}{10}+1\)

=>\(\frac56<\frac78<\frac{9}{10}\) <1

mà \(1=\frac88;1=\frac55<\frac65\)

nên \(\frac56<\frac78<\frac{9}{10}<\frac88<\frac65\)

\(a.\frac{3}{7}.\frac{5}{6}+\frac{5}{6}.\frac{4}{7}=\left(\frac{3}{7}+\frac{4}{7}\right).\frac{5}{6}=1.\frac{5}{6}=\frac{5}{6}\)

\(b.\frac{1}{6}:\frac{4}{5}+\frac{1}{8}:\frac{4}{5}=\left(\frac{1}{6}+\frac{1}{8}\right):\frac{4}{5}=\left(\frac{4}{24}+\frac{3}{24}\right):\frac{4}{5}=\frac{7}{24}:\frac{4}{5}=\frac{7.5}{24.4}\)\(=\frac{35}{96}\)

\(c.\frac{1}{6}:\frac{4}{5}-\frac{1}{8}:\frac{4}{5}=\left(\frac{1}{6}-\frac{1}{8}\right):\frac{4}{5}=\left(\frac{4}{24}-\frac{3}{24}\right):\frac{4}{5}\)\(=\frac{1}{24}:\frac{4}{5}=\frac{1.5}{24.4}=\frac{5}{96}\)

a) \(\frac{3}{7}\cdot\frac{5}{6}+\frac{5}{6}\cdot\frac{4}{7}\)

\(=\left(\frac{3}{7}+\frac{4}{7}\right)\cdot\frac{5}{6}\)

\(=1\cdot\frac{5}{6}\)

\(=\frac{5}{6}\)

b) \(\frac{1}{6}:\frac{4}{5}+\frac{1}{8}:\frac{4}{5}\)

\(=\left(\frac{1}{6}+\frac{1}{8}\right):\frac{4}{5}\)

\(=\frac{7}{24}:\frac{4}{5}\)

\(=\frac{35}{96}\)

c) \(\frac{1}{6}:\frac{4}{5}-\frac{1}{8}:\frac{4}{5}\)

\(=\left(\frac{1}{6}-\frac{1}{8}\right):\frac{4}{5}\)

\(=\frac{1}{24}:\frac{4}{5}\)

\(=\frac{5}{96}\)

\(\frac{1}{2}:\frac{3}{2}:\frac{5}{4}:\frac{6}{5}:\frac{7}{6}:\frac{8}{7}\)

\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\)

\(=\frac{1\cdot\left(2\cdot5\cdot6\cdot7\right)}{8\cdot3\cdot\left(2\cdot5\cdot6\cdot7\right)}\)

\(=\frac{1}{24}\)

\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot\frac{5}{6}\cdot\frac{6}{7}\cdot\frac{7}{8}\cdot\frac{8}{9}\cdot\frac{9}{10}\)

\(=\frac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)}{\left(2\cdot3\cdot4\cdot5\cdot6\cdot7\cdot8\cdot9\right)\cdot10}\)

\(=\frac{1}{10}\)

\(1.a,\frac{5}{7}=\frac{5.9}{7.9}=\frac{45}{63};\frac{4}{9}=\frac{4.7}{9.7}=\frac{28}{63}.\)

\(b,\frac{7}{15},\frac{5}{3}=\frac{5.5}{3.5}=\frac{25}{15}\)

\(c,\frac{11}{12}=\frac{11.4}{12.4}=\frac{44}{48};\frac{7}{48}\)

\(d,\frac{3}{2}=\frac{3.3}{2.3}=\frac{9}{6};\frac{2}{3}=\frac{2.2}{3.2}=\frac{4}{6}\)

\(e,\frac{1}{3}=\frac{1.4}{3.4}=\frac{4}{12};\frac{5}{4}=\frac{5.3}{4.3}=\frac{15}{12};\frac{10}{12}\)

a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)

\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)

b)\(5+\dfrac{3}{5}=\dfrac{25}{5}+\dfrac{3}{5}=\dfrac{28}{5};10-\dfrac{9}{16}=\dfrac{160}{16}-\dfrac{9}{16}=\dfrac{151}{16}\)

\(\dfrac{2}{3}-\left(\dfrac{1}{6}+\dfrac{1}{8}\right)=\dfrac{2}{3}-\left(\dfrac{8}{48}+\dfrac{6}{48}\right)=\dfrac{2}{3}-\dfrac{14}{48}=\dfrac{32}{48}-\dfrac{14}{48}=\dfrac{3}{8}\)

Gợi ý: Rút gọn 2 ps, quy đồng rồi so sánh.

15/50 ? 90/45 ra 0.26<2