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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)
\({\left( {0,2} \right)^2}.{\left( {0,2} \right)^3} = \left( {0,2.0,2} \right).\left( {0,2.0,2.0,2} \right) = {\left( {0,2} \right)^5}\)
a)
\(\frac{2}{5} = \frac{{40}}{{100}}\)
\( - \frac{3}{4} = - \frac{{75}}{{100}};\)
\(\frac{{ - 1,3}}{{10}} = - 13.\)
b) Khối lượng nước trong cơ thể Hùng là:
\(40.\frac{{70}}{{100}} = 28\,\,(kg)\)
\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)+\left(\frac{-2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}\right)-\frac{5}{2}\)
\(=-2+0+2-\frac{5}{2}=\frac{-5}{2}\)
\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)+\left(\frac{7}{3}-\frac{2}{3}-\frac{5}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2+0-\frac{1}{2}=-\frac{5}{2}\)
\(A=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)
\(A=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{5}{3}-\frac{1}{3}-\frac{4}{3}\right)\)
\(A=-16+\frac{1}{4}+0\)
\(A=-15\frac{3}{4}\)
\(A=\left(-7+\frac{3}{4}-\frac{1}{3}\right)-\left(6-\frac{5}{4}+\frac{4}{3}\right)-\left(3+\frac{7}{4}-\frac{5}{3}\right)\)
\(=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)
\(=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{-1}{3}-\frac{4}{3}+\frac{5}{3}\right)\)
\(=-16-\frac{1}{4}\)
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= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
= \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
= \(\left(6-5-3\right)\)\(+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)\)\(+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
= \(-2+0+\left(\frac{-1}{2}\right)\)=\(\frac{-5}{2}\)
Cách 1:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{30}{6}+\frac{10}{6}-\frac{9}{6}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)
\(=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
\(=-\frac{15}{6}\)
\(=-\frac{5}{2}\)
Cách 2:
\(A=\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)
\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2+0-\frac{1}{2}\)
\(=-\frac{4}{2}-\frac{1}{2}\)
\(=-\frac{5}{2}\)
C1: \(A=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}\)
\(A=\frac{35-31-19}{6}=\frac{-15}{6}=\frac{-5}{2}\)
C2: \(A=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(A=\left(6-5-3\right)+\left(-\frac{2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(A=-2-\frac{1}{2}=\frac{-5}{2}\)
C1:
\(A=\left(\frac{36-4+3}{6}\right)-\left(\frac{30+10-9}{6}\right)-\left(\frac{18-14+15}{6}\right)\)
\(A=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{-15}{6}=\frac{-5}{2}\)
C2: \(A=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(A=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(A=-2-0+\frac{-1}{2}=\frac{-5}{2}\)
Trên đây là bài giải của mình, bạn hãy tính lại bằng máy tính xem Kq có đúng ko nhé!
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.