Bộ 10 đề thi cuối kì 1 Toán 7 Kết nối tri thức có đáp án - Đề 4

Thực hiện phép tính (tính hợp lí nếu có thể): a) 2/9 − 5/7 . 14/3 ; b) 3/2 . 1/5 + 3/2 . 4/5 − 7/4 ;

9/14

PHẦN II. TỰ LUẬN (8,0 điểm)

(2,0 điểm) Thực hiện phép tính (tính hợp lí nếu có thể):

a) \(\frac{2}{9} - \frac{5}{7}\,\,.\,\,\frac{{14}}{3}\);                                                             

b) \[\frac{3}{2}\,\,.\,\,\frac{1}{5} + \frac{3}{2}\,\,.\,\,\frac{4}{5} - \frac{7}{4}\];

c) \[{\left( {\frac{{ - 4}}{7}} \right)^2}:{\left( {\frac{{ - 2}}{7}} \right)^2} + 6\,\,.\,\,\left( {\frac{{ - 1}}{6}} \right) - \sqrt {\frac{4}{{25}}} \];                              

d) \(\sqrt {\frac{1}{{25}}} \,\,.\,\,{( - 2)^2} - \sqrt {\frac{1}{{81}}} :{\left( {\frac{{ - 2}}{3}} \right)^3}\).

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Giải thích

a) \(\frac{2}{9} - \frac{5}{7}\,\,.\,\,\frac{{14}}{3} = \frac{2}{9} - 5\,\,.\,\,\frac{2}{3} = \frac{2}{9} - \frac{{10}}{3} = \frac{{ - 28}}{9}\);

b) \[\frac{3}{2}\,\,.\,\,\frac{1}{5} + \frac{3}{2}\,\,.\,\,\frac{4}{5} - \frac{7}{4} = \frac{3}{2}\,\,.\,\,\left( {\frac{1}{5} + \frac{4}{5}} \right) - \frac{7}{4}\]

\[ = \frac{3}{2}\,\,.\,\,1 - \frac{7}{4} = \frac{3}{2}\, - \frac{7}{4} = \frac{{ - 1}}{4}\];

c) \[{\left( {\frac{{ - 4}}{7}} \right)^2}:{\left( {\frac{{ - 2}}{7}} \right)^2} + 6\,\,.\,\,\left( {\frac{{ - 1}}{6}} \right) - \sqrt {\frac{4}{{25}}} \]

\[ = \frac{{{{\left( { - 4} \right)}^2}}}{{{7^2}}}:\frac{{{{\left( { - 2} \right)}^2}}}{{{7^2}}} + \left( { - 1} \right) - \frac{2}{5}\]

\[ = \frac{{{4^2}}}{{{7^2}}}\,\,.\,\,\frac{{{7^2}}}{{{2^2}}} - 1 - \frac{2}{5} = 4 - 1 - \frac{2}{5} = 3 - \frac{2}{5} = \frac{{13}}{5}\];                       

d) \[\sqrt {\frac{1}{{25}}} \,\,.\,\,{( - 2)^2} - \sqrt {\frac{1}{{81}}} :{\left( {\frac{{ - 2}}{3}} \right)^3} = \frac{1}{5}\,\,.\,\,4 - \frac{1}{9}:\frac{{{{\left( { - 2} \right)}^3}}}{{{3^3}}}\]

\[ = \frac{4}{5} - \frac{1}{{{3^2}}}\,\,.\,\,\frac{{{3^3}}}{{{{\left( { - 2} \right)}^3}}} = \frac{4}{5} + \frac{3}{8} = \frac{{47}}{{40}}\].