Ta có: \(\dfrac{2}{4} = \dfrac{3}{6} = \dfrac{1}{2}\)
Lại có:
\(\eqalign{& {{2 + 3} \over {4 + 6}} = {5 \over {10}} = {1 \over 2} \cr & {{2 - 3} \over {4 - 6}} = {{ - 1} \over { - 2}} = {1 \over 2} \cr & \Rightarrow {2 \over 4} = {3 \over 6} = {{2 + 3} \over {4 + 6}} = {{2 - 3} \over {4 - 6}} \cr} \)
