Bộ 5 đề thi giữa kì 1 Toán 7 Cánh diều cấu trúc mới có đáp án - Đề 3

(1,5 điểm) Thực hiện phép tính: a) 31/23 − ( 7/32 + 8/23 ) ; b) ( 1 − 2 3 − 1 4 ) : ( 4 5 − 3 4 ) 2 ; c) ( − 1 2 ) 2 . 2 3 + √ 81 . ( − 2 3 ) + 1 1 2 . √ 4 9 .

19/21

(1,5 điểm) Thực hiện phép tính:

a) \(\frac{{31}}{{23}} - \left( {\frac{7}{{32}} + \frac{8}{{23}}} \right)\);

b) \(\left( {1 - \frac{2}{3} - \frac{1}{4}} \right):{\left( {\frac{4}{5} - \frac{3}{4}} \right)^2}\);

c) \({\left( { - \frac{1}{2}} \right)^2}.\frac{2}{3} + \sqrt {81} .\left( {\frac{{ - 2}}{3}} \right) + 1\frac{1}{2}.\sqrt {\frac{4}{9}} \).

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Hướng dẫn giải

a) \(\frac{{31}}{{23}} - \left( {\frac{7}{{32}} + \frac{8}{{23}}} \right)\)

\( = \frac{{31}}{{23}} - \frac{7}{{32}} - \frac{8}{{23}}\)

\( = \left( {\frac{{31}}{{23}} - \frac{8}{{23}}} \right) - \frac{7}{{32}}\)

\( = \frac{{23}}{{23}} - \frac{7}{{32}}\)

\( = 1 - \frac{7}{{32}}\)

\( = \frac{{25}}{{32}}.\)

b) \(\left( {1 - \frac{2}{3} - \frac{1}{4}} \right):{\left( {\frac{4}{5} - \frac{3}{4}} \right)^2}\)

\( = \left( {\frac{{12}}{{12}} - \frac{8}{{12}} - \frac{3}{{12}}} \right):{\left( {\frac{{16}}{{20}} - \frac{{15}}{{20}}} \right)^2}\)

\( = \frac{1}{{12}}:{\left( {\frac{1}{{20}}} \right)^2}\)

\( = \frac{1}{{12}}:\frac{1}{{400}}\)

\( = \frac{1}{{12}}.400\)

\( = \frac{{400}}{{12}}\)

\( = \frac{{100}}{3}.\)

c) \({\left( { - \frac{1}{2}} \right)^2}.\frac{2}{3} + \sqrt {81} .\left( {\frac{{ - 2}}{3}} \right) + 1\frac{1}{2}.\sqrt {\frac{4}{9}} \)

\( = \frac{1}{4}.\frac{2}{3} - \sqrt {{9^2}} .\frac{2}{3} + \frac{3}{2}.\sqrt {{{\left( {\frac{2}{3}} \right)}^2}} \)

\( = \frac{1}{4}.\frac{2}{3} - 9.\frac{2}{3} + \frac{3}{2}.\frac{2}{3}\)

\( = \left( {\frac{1}{4} - 9 + \frac{3}{2}} \right).\frac{2}{3}\)

\( = \left( {\frac{1}{4} - \frac{{36}}{4} + \frac{6}{4}} \right).\frac{2}{3}\)

\( = - \frac{{29}}{4}.\frac{2}{3}\)

\( = - \frac{{29}}{6}\).