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:

\(40.\frac{{70}}{{100}} = 28\,\,(kg)\)

\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)+\left(\frac{-2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}\right)-\frac{5}{2}\)

\(=-2+0+2-\frac{5}{2}=\frac{-5}{2}\)

\(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

\(=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)+\left(\frac{7}{3}-\frac{2}{3}-\frac{5}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)

\(=-2+0-\frac{1}{2}=-\frac{5}{2}\)

\(A=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)

\(A=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{5}{3}-\frac{1}{3}-\frac{4}{3}\right)\)

\(A=-16+\frac{1}{4}+0\)

\(A=-15\frac{3}{4}\)

\(A=\left(-7+\frac{3}{4}-\frac{1}{3}\right)-\left(6-\frac{5}{4}+\frac{4}{3}\right)-\left(3+\frac{7}{4}-\frac{5}{3}\right)\)

\(=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)

\(=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{-1}{3}-\frac{4}{3}+\frac{5}{3}\right)\)

\(=-16-\frac{1}{4}\)

#)Giải :

\(A=\left(-7+\frac{3}{4}-\frac{1}{3}\right)-\left(6-\frac{5}{4}+\frac{4}{3}\right)-\left(3+\frac{7}{4}-\frac{5}{3}\right)\)

\(A=-7+\frac{3}{4}-\frac{1}{3}-6+\frac{5}{4}-\frac{4}{3}-3-\frac{7}{4}+\frac{5}{3}\)

\(A=\left(-7-6-3\right)+\left(\frac{3}{4}+\frac{5}{4}-\frac{7}{4}\right)+\left(\frac{1}{3}-\frac{4}{3}+\frac{5}{3}\right)\)

\(A=\left(-16\right)+\frac{1}{4}+\frac{2}{3}\)

\(A=\frac{-179}{12}\)

\(B=\left(6+\frac{2}{3}-\frac{1}{2}\right)-\left(5-\frac{5}{3}+\frac{3}{2}\right)-\left(3+\frac{7}{3}-\frac{5}{2}\right)\)

\(B=6+\frac{2}{3}-\frac{1}{2}-5+\frac{5}{3}-\frac{3}{2}-3-\frac{7}{3}+\frac{5}{2}\)

\(B=\left(6-5-3\right)+\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}-\frac{3}{2}+\frac{5}{2}\right)\)

\(B=\left(-2\right)+0+\frac{3}{2}\)

\(B=\frac{-1}{2}\)

\(A=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(=\left(6-5-3\right)+\left(\frac{3}{2}+\frac{1}{2}-\frac{5}{2}\right)+\left(\frac{7}{3}-\frac{2}{3}-\frac{5}{3}\right)\)

= -2 + \(\frac{-1}{2}\)+ 0 = \(\frac{-5}{2}\)

Cách 1:  A= \(\left(6-\frac{2}{3}+\frac{1}{2}\right)-\left(5+\frac{5}{3}-\frac{3}{2}\right)\)\(-\left(3-\frac{7}{3}+\frac{5}{2}\right)\)

               =  \(\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=\frac{-15}{6}\)=\(\frac{-5}{2}\)

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
23+12)−(5+53−32)−(3−73+52)=(6.66−2.26+36)−(5.66+5.26−3.36)−(3.66−7.26+5.36)=36−4+36−30+10−96−18−14+156=356−316−196=35−31−196=−156=−52=−212.

23+12)−(5+53−32)−(3−73+52)=6−23+12−5−53+32−3+73−52=(6−5−3)+(−23−53+73)+(12+32−52)=−2+−2−5+73+1+3−52=−2+0−12=−52=−212." role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.94px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; overflow-wrap:normal; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal" class="MathJax_CHTML mjx-chtml">A=(6−23+12)−(5+53−32)−(3−73+52)=6−23+12−5−53+32−3+73−52=(6−5−3)+(−23−53+73)+(12+32−52)=−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: \(A=6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)

\(A=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)

\(A=-2-0+\frac{-1}{2}=\frac{-5}{2}\)

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