Giải Bài 34 trang 22 sách bài tập toán 7 tập 1 - Cánh diều

Đề bài

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

Lời giải chi tiết

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