thực hiện phép tính:
a)\(\frac{3}{2}+\left(6-\frac{3}{2}\right)\)
b)\(\frac{7}{3}-\frac{4}{3}:4+1\frac{1}{2}\)
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Những câu hỏi liên quan
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\= - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)
b)
\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)
c)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)
d)
\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ = - \frac{3}{4}\end{array}\).
22 tháng 10 2017
\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
\(=3-\frac{1}{4}+\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
\(=\left(3-5-6\right)+\left(-\frac{1}{4}+\frac{7}{4}-\frac{3}{2}\right)+\left(\frac{2}{3}-\frac{1}{3}\right)+\frac{6}{5}\)
\(=-8+\frac{1}{3}+\frac{6}{5}\)
\(=-\frac{97}{15}\)
22 tháng 10 2017
= 3 - 1/4 +2/3 - 5 - 1/3 + 6/5 - 6 + 7/4 - 3/2
= 2/3 . -3/2 . ( 3 + 5 + 6 ) . ( 2/3 + 1/3 ) . ( -1/4 - 7/4)
= -1 . 14 . 1 . 6/4
= -14 . 1 . 6/4
= -14 . 6/4
= -84/4 = -21
15 tháng 3 2020
Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\)
=\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\)
\(=\frac{5}{2}.\frac{6}{5}-17\)
\(=3-17=-14\)
Tụi quá mới lớp 5 thui
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)
b)
\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ = - 2 - \frac{5}{7}.\frac{7}{{15}}\\ = - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)
c)
\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)
d)
\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)
B
30 tháng 4 2019
\(\left(\frac{3}{8}+-\frac{3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\left(\frac{9}{24}+-\frac{18}{24}+\frac{14}{24}\right):\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}:\frac{5}{6}+\frac{1}{2}\)
\(=\frac{5}{24}.\frac{6}{5}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{1}{2}\)
\(=\frac{1}{4}+\frac{2}{4}\)
\(=\frac{3}{4}\)
B
30 tháng 4 2019
\(\frac{1}{2}+\frac{3}{4}-\left(\frac{3}{4}-\frac{4}{5}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\left(\frac{15}{20}-\frac{16}{20}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{-1}{20}\)
\(=\frac{10}{20}+\frac{15}{20}-\frac{-1}{20}\)
\(=\frac{25}{20}-\frac{-1}{20}\)
\(=\frac{26}{20}\)
\(=\frac{13}{10}\)
HQ
Hà Quang Minh
Giáo viên
18 tháng 9 2023
a)
\(\begin{array}{l}{\left( {1 + \frac{1}{2} - \frac{1}{4}} \right)^2}.\left( {2 + \frac{3}{7}} \right)\\ = {\left( {\frac{4}{4} + \frac{2}{4} - \frac{1}{4}} \right)^2}.\left( {\frac{{14}}{7} + \frac{3}{7}} \right)\\ = {\left( {\frac{5}{4}} \right)^2}.\frac{{17}}{7}\\ = \frac{{25}}{{16}}.\frac{{17}}{7}\\ = \frac{{425}}{{112}}\end{array}\)
b)
\(\begin{array}{l}4:{\left( {\frac{1}{2} - \frac{1}{3}} \right)^3}\\ = 4:{\left( {\frac{3}{6} - \frac{2}{6}} \right)^3}\\ = 4:{\left( {\frac{1}{6}} \right)^3}\\ = 4:\frac{1}{{216}}\\ = 4.216\\ = 864\end{array}\)
a) \(\frac{3}{2}+\left(6-\frac{3}{2}\right)\)
\(=\frac{3}{2}+6-\frac{3}{2}\)
\(=\left(\frac{3}{2}-\frac{3}{2}\right)+6\)
\(=0+6\)
\(=6\)
b) \(\frac{7}{3}-\frac{4}{3}:4+1\frac{1}{2}\)
\(=\frac{7}{3}-\frac{1}{3}+\frac{3}{2}\)
\(=2+\frac{3}{2}\)
\(=\frac{7}{2}\)
_Chúc bạn học tốt_
a) \(\frac{3}{2}+\left(6-\frac{3}{2}\right)\)
\(=\frac{3}{2}+6-\frac{3}{2}\)
\(=\frac{3}{2}-\frac{3}{2}+6\)
\(=0+6=6\)
Đây là cách tính nhanh nha
\(\frac{7}{3}-\frac{4}{3}:4+1\frac{1}{2}\)
\(=\frac{7}{3}-\frac{4}{3}.\frac{1}{4}+\frac{3}{2}\)
\(=\frac{7}{3}-\frac{1}{3}+\frac{3}{2}\)
\(=\frac{6}{3}+\frac{3}{2}\)
\(=\frac{12+9}{6}\)
\(=\frac{21}{6}\)
\(=\frac{7}{2}\)