\(\dfrac{-5}{7}=\dfrac{-5\cdot19}{7\cdot19}=\dfrac{-95}{133}\)
\(\dfrac{-16}{19}=\dfrac{-16\cdot7}{19\cdot7}=\dfrac{-112}{133}\)
mà -95>-112
nên \(-\dfrac{16}{19}< -\dfrac{5}{7}\)(1)
\(-\dfrac{5}{7}< 0\)
\(0< \dfrac{214}{315}\)
Do đó: \(-\dfrac{5}{7}< \dfrac{214}{315}\left(3\right)\)
\(\dfrac{214}{315}< \dfrac{220.5}{315}=\dfrac{7}{10}\)(2)
\(\dfrac{7}{10}=1-\dfrac{3}{10}\)
\(\dfrac{20}{23}=1-\dfrac{3}{23}\)
ta có: \(\dfrac{3}{10}>\dfrac{3}{23}\)
=>\(-\dfrac{3}{10}< -\dfrac{3}{23}\)
=>\(-\dfrac{3}{10}+1< -\dfrac{3}{23}+1\)
=>\(\dfrac{7}{10}< \dfrac{20}{23}\)(4)
\(\dfrac{20}{23}< 1\)
\(1< \dfrac{205}{107}\)
DO đó: \(\dfrac{20}{23}< \dfrac{205}{107}\)(5)
Từ (1),(2),(3),(4),(5) suy ra \(\dfrac{-16}{19}< -\dfrac{5}{7}< \dfrac{214}{315}< \dfrac{7}{10}< \dfrac{20}{23}< \dfrac{205}{107}\)