Tìm số thích hợp thay vào dấu “?” trong các câu dưới đây:
a)\({\left( {\frac{1}{3}} \right)^2}.{\left( {\frac{1}{3}} \right)^2} = {\left( {\frac{1}{3}} \right)^?}\) b)\({\left( {0,2} \right)^2}.{\left( {0,2} \right)^3} = {\left( {0,2} \right)^?}\).
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HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)\({\left[ {{{\left( {\frac{{ - 2}}{3}} \right)}^2}} \right]^5} = {\left( {\frac{{ - 2}}{3}} \right)^{2.5}} = {\left( {\frac{{ - 2}}{3}} \right)^{10}}\)
Vậy dấu “?” bằng 10.
b) \({\left[ {{{\left( {0,4} \right)}^3}} \right]^3} = {\left( {0,4} \right)^{3.3}} = {\left( {0,4} \right)^9}\)
Vậy dấu “?” bằng 9.
c) \({\left[ {{{\left( {7,31} \right)}^3}} \right]^0} = 1\)
Vậy dấu “?” bằng 1.
17 tháng 6 2019
\(A=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)
\(A=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{5}{3}-\frac{1}{3}-\frac{4}{3}\right)\)
\(A=-16+\frac{1}{4}+0\)
\(A=-15\frac{3}{4}\)
17 tháng 6 2019
\(A=\left(-7+\frac{3}{4}-\frac{1}{3}\right)-\left(6-\frac{5}{4}+\frac{4}{3}\right)-\left(3+\frac{7}{4}-\frac{5}{3}\right)\)
\(=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)
\(=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{-1}{3}-\frac{4}{3}+\frac{5}{3}\right)\)
\(=-16-\frac{1}{4}\)
S
3 tháng 7 2017
\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)+\left(\frac{-2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}\right)-\frac{5}{2}\)
\(=-2+0+2-\frac{5}{2}=\frac{-5}{2}\)
3 tháng 7 2017
\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)+\left(\frac{7}{3}-\frac{2}{3}-\frac{5}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2+0-\frac{1}{2}=-\frac{5}{2}\)
2 tháng 8 2019
\(\left(-2\right).\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right).....\left(-1\frac{1}{2013}\right)\)
\(=\left(-2\right).\left(\frac{-3}{2}\right)\left(-\frac{4}{3}\right)......\left(\frac{-2014}{2013}\right)\)
\(=\frac{\left(-2\right).\left(-3\right).\left(-4\right)....\left(-2014\right)}{2.3.....2013}\)
\(=\frac{2.3.4....2014\left(\text{Vì có 2014 thừa số âm }\right)}{2.3....2013}\)
\(=\frac{\left(2.3.4....2013\right).2014}{2.3....2013}\)
\(=2014\)
18 tháng 12 2016
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
24 tháng 6 2016
Cách 1: A= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{-15}{6}\)=\(\frac{-5}{2}\)
Cách 2: A= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
= \(\left(6-5-3\right)\)\(+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)\)\(+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
= \(-2+0+\left(\frac{-1}{2}\right)\)=\(\frac{-5}{2}\)
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)
\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ = - \frac{3}{4}\end{array}\).
a) Ta có:
\({\left( {\frac{1}{3}} \right)^2}.{\left( {\frac{1}{3}} \right)^2} = \frac{1}{3}.\frac{1}{3}.\frac{1}{3}\frac{1}{3} = {\left( {\frac{1}{3}} \right)^4}\)
b)
\({\left( {0,2} \right)^2}.{\left( {0,2} \right)^3} = \left( {0,2.0,2} \right).\left( {0,2.0,2.0,2} \right) = {\left( {0,2} \right)^5}\)