a) $\frac{2}{5} < \frac{3}{5}$ ; $\frac{4}{7} > \frac{2}{7}$ ; $\frac{1}{{10}} < \frac{7}{{10}}$ ; $\frac{5}{2} > \frac{2}{2}$
b) $\frac{1}{3}$ và $\frac{5}{{12}}$
$\frac{1}{3} = \frac{{1 \times 4}}{{3 \times 4}} = \frac{4}{{12}}$
Ta có $\frac{4}{{12}} < \frac{5}{{12}}$ vậy $\frac{1}{3} < \frac{5}{{12}}$
$\frac{1}{5}$ và $\frac{2}{{15}}$
$\frac{1}{5} = \frac{{1 \times 3}}{{5 \times 3}} = \frac{3}{{15}}$
Ta có $\frac{3}{{15}} > \frac{2}{{15}}$ vậy $\frac{1}{5} > \frac{2}{{15}}$
$\frac{5}{{14}}$ và $\frac{1}{2}$
$\frac{1}{2} = \frac{{1 \times 7}}{{2 \times 7}} = \frac{7}{{14}}$
Ta có $\frac{5}{{14}} < \frac{7}{{14}}$ vậy $\frac{5}{{14}} < \frac{1}{2}$
$\frac{2}{3}$ và $\frac{{12}}{{18}}$
$\frac{2}{3} = \frac{{2 \times 6}}{{3 \times 6}} = \frac{{12}}{{18}}$
Ta có $\frac{{12}}{{18}} = \frac{{12}}{{18}}$ vậy $\frac{2}{3} = \frac{{12}}{{18}}$
a: 2<3
nên \(\dfrac{2}{5}< \dfrac{3}{5}\)
4>2
nên \(\dfrac{4}{7}>\dfrac{2}{7}\)
1<7
nên \(\dfrac{1}{10}< \dfrac{7}{10}\)
5>2
nên \(\dfrac{5}{2}>\dfrac{2}{2}\)
b: \(\dfrac{1}{3}=\dfrac{1\cdot4}{3\cdot4}=\dfrac{4}{12}< \dfrac{5}{12}\)
\(\dfrac{1}{5}=\dfrac{1\cdot3}{3\cdot5}=\dfrac{3}{15}>\dfrac{2}{15}\)
\(\dfrac{1}{2}=\dfrac{1\cdot7}{2\cdot7}=\dfrac{7}{14}>\dfrac{5}{14}\)
\(\dfrac{2}{3}=\dfrac{2\cdot6}{3\cdot6}=\dfrac{12}{18}\)
