Điền dấu thích hợp (>; <; =) vào chỗ chấm :
\(\dfrac{5}{8}....\dfrac{8}{{11}}\) \(\dfrac{7}{9}....\dfrac{5}{8}\)
\(\dfrac{{12}}{{18}}....\dfrac{{18}}{{27}}\) \(\dfrac{{25}}{{36}}....\dfrac{{25}}{{63}}\)
• Ta có :
\(\dfrac{5}{8} = \dfrac{{5 \times 11}}{{8 \times 11}} = \dfrac{{55}}{{88}}\;;\) \(\dfrac{8}{{11}} = \dfrac{{8 \times 8}}{{11 \times 8}} = \dfrac{{64}}{{88}}\)
Mà : \( \dfrac{{55}}{{88}}< \dfrac{{64}}{{88}}\). Vậy \(\dfrac{5}{8} < \dfrac{8}{{11}}\).
• Ta có :
\(\dfrac{7}{9} = \dfrac{{ 7\times 8}}{{ 9\times 8}} = \dfrac{{56}}{{72}}\;;\) \(\dfrac{5}{{8}} = \dfrac{{ 5\times 9}}{{ 8\times 9}} = \dfrac{{45}}{{72}}\)
Mà : \( \dfrac{{56}}{{72}}> \dfrac{{45}}{{72}}\). Vậy \(\dfrac{7}{9} > \dfrac{5}{{8}}\).
• Ta có :
\(\dfrac{12}{18} = \dfrac{{ 12:6 }}{{ 18:6 }} = \dfrac{{2}}{{3}}\;;\) \(\dfrac{18}{{27}} = \dfrac{{ 18:9 }}{{ 27:9 }} = \dfrac{{2}}{{3}}\)
Mà : \( \dfrac{{2}}{{3}}=\dfrac{{2}}{{3}}\). Vậy \(\dfrac{12}{18} = \dfrac{18}{{27}}\).
• Ta có : \( 36 > 63\). Vậy \(\dfrac{25}{36} >\dfrac{25}{{63}}\).
-- Mod Toán lớp 4