Bài 34 trang 22 SBT Toán lớp 7 tập 1 Cánh diều: Chọn dấu “<”, “>”, “=” thích hợp cho :...

Chọn dấu “<”, “>”, “=” thích hợp cho :

a) $\dfrac{5}{6} - {\left( {\dfrac{1}{6}} \right)^2}$ ${\left( {\dfrac{5}{6} - \dfrac{1}{6}} \right)^2}$;

b) $250.{\left( {\dfrac{1}{5} - \dfrac{1}{6}} \right)^2}$$250.{\left( {\dfrac{1}{5}} \right)^2} - \dfrac{1}{6}$;

c) $3\dfrac{1}{5}:1,5 + 4\dfrac{2}{5}:1,5$$\left( {3\dfrac{1}{5} + 4\dfrac{2}{5}} \right):1,5$;

d) $\left( {\dfrac{9}{{25}} - 2,18} \right):\left( {3\dfrac{4}{5} + 0,2} \right)$$\dfrac{9}{{25}}:3\dfrac{4}{5} - 2,18:0,2$.

Muốn điền được dấu của các phần, ta thực hiện các phép tính ở hai vế rồi so sánh chúng với nhau.

a) Ta có:

     $\dfrac{5}{6} - {\left( {\dfrac{1}{6}} \right)^2} = \dfrac{5}{6} - \dfrac{1}{{36}} \\= \dfrac{{30}}{{36}} - \dfrac{1}{{36}} = \dfrac{{29}}{{36}}$

     ${\left( {\dfrac{5}{6} - \dfrac{1}{6}} \right)^2} = {\left( {\dfrac{4}{6}} \right)^2}\\ = \dfrac{{16}}{{36}}$

Vì 29 > 16 nên $\dfrac{{29}}{{36}} > \dfrac{{16}}{{36}}$ nên: $\dfrac{5}{6} - {\left( {\dfrac{1}{6}} \right)^2}$> ${\left( {\dfrac{5}{6} - \dfrac{1}{6}} \right)^2}$.

b) Ta có:

     $250.{\left( {\dfrac{1}{5} - \dfrac{1}{6}} \right)^2} = 250.{\left( {\dfrac{6}{{30}} - \dfrac{5}{{30}}} \right)^2}\\ = 250.{\left( {\dfrac{1}{{30}}} \right)^2} = 250.\dfrac{1}{{900}} \\= \dfrac{{250}}{{900}} = \dfrac{5}{{18}}$

     $250.{\left( {\dfrac{1}{5}} \right)^2} - \dfrac{1}{6} = 250.\dfrac{1}{{25}} - \dfrac{1}{6} \\= 10 - \dfrac{1}{6} = \dfrac{{60}}{6} - \dfrac{1}{6}\\ = \dfrac{{59}}{6} = \dfrac{{177}}{{18}}$

Vì 5 < 177 nên $\dfrac{5}{{18}} < \dfrac{{177}}{{18}}$ nên: $250.{\left( {\dfrac{1}{5} - \dfrac{1}{6}} \right)^2}$< $250.{\left( {\dfrac{1}{5}} \right)^2} - \dfrac{1}{6}$.

c) Ta có:

     $3\dfrac{1}{5}:1,5 + 4\dfrac{2}{5}:1,5 = \left( {3\dfrac{1}{5} + 4\dfrac{2}{5}} \right):1,5$

Vậy  $3\dfrac{1}{5}:1,5 + 4\dfrac{2}{5}:1,5=\left( {3\dfrac{1}{5} + 4\dfrac{2}{5}} \right):1,5$

d) Ta có:

     $\left( {\dfrac{9}{{25}} - 2,18} \right):\left( {3\dfrac{4}{5} + 0,2} \right) \\= \left( {\dfrac{9}{{25}} - \dfrac{{218}}{{100}}} \right):\left( {\dfrac{{19}}{5} + \dfrac{2}{{10}}} \right)\\ = \left( {\dfrac{{ - 91}}{{50}}} \right):\dfrac{{40}}{{10}} = \dfrac{{ - 91}}{{200}}$

     $\dfrac{9}{{25}}:3\dfrac{4}{5} - 2,18:0,2\\ = \dfrac{9}{{25}}:\dfrac{{19}}{5} - \dfrac{{218}}{{100}}:\dfrac{2}{{10}}\\ = \dfrac{9}{{25}}.\dfrac{5}{{19}} - \dfrac{{218}}{{100}}.\dfrac{{10}}{2}\\ = \dfrac{9}{{95}} - \dfrac{{109}}{{10}} \\= \dfrac{18}{{190}} - \dfrac{{2071}}{{10}} \\= \dfrac{{ - 2053}}{{190}}$

Vì $(-91) > (-2053)$ nên $\dfrac{{ - 91}}{{200}} > \dfrac{{ - 2053}}{{190}}$ nên: $\left( {\dfrac{9}{{25}} - 2,18} \right):\left( {3\dfrac{4}{5} + 0,2} \right)$> $\dfrac{9}{{25}}:3\dfrac{4}{5} - 2,18:0,2$.