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) $\frac{{14}}{{18}}:\frac{8}{9} = \frac{7}{9}:\frac{8}{9} = \frac{7}{9} \times \frac{9}{8} = \frac{{63}}{{72}} = \frac{7}{8}$

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) Thứ tự từ bé đến lớn: \(\dfrac{2}{7};\dfrac{3}{7};\dfrac{5}{7}\)

b) Thứ tự từ bé đến lớn: \(\dfrac{1}{8};\dfrac{5}{8};\dfrac{7}{8}\)

c) Thứ tự từ bé đến lớn: \(\dfrac{1}{10};\dfrac{7}{10};\dfrac{9}{10}\)

a) \(\dfrac{2}{5}=\dfrac{2\times5}{5\times2}=\dfrac{10}{25}\)

\(\dfrac{4}{7}=\dfrac{4\times2}{7\times2}=\dfrac{8}{14}\)

b) \(\dfrac{36}{40}=\dfrac{36:4}{40:4}=\dfrac{9}{10}\)

\(\dfrac{24}{32}=\dfrac{24:8}{32:8}=\dfrac{3}{4}\)

a) \(< \)

b) \(>\)

c) \(< \)

d) \(>\)

e) \(< \)

g) \(>\)

h) \(>\)

k) \(>\)

=9/10-2/10+1

=7/10+10/10

=17/10

17/10

gì zậy

j vậy error à?

\(\dfrac{2}{3}:\dfrac{7}{5}:\dfrac{x}{9}=\dfrac{2}{7}x\dfrac{3}{5}x\dfrac{9}{10}\)

\(\dfrac{2}{3}:\dfrac{7}{5}:\dfrac{x}{9}=\dfrac{3}{7}\)

\(\dfrac{10}{21}:\dfrac{x}{9}=\dfrac{3}{7}\)

\(\dfrac{x}{9}=\dfrac{10}{21}:\dfrac{3}{7}\)

\(\dfrac{x}{9}=\dfrac{10}{9}\)

\(=>x=10\)

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?