e) \(E=0,7.2\frac{2}{3}.20.0,375.\frac{5}{28}\)

        \(=\left(0,7.20\right)\left(2\frac{2}{3}.0,375\right)\frac{5}{28}\)

        \(=14.1.\frac{5}{28}\)

        \(=14.\frac{5}{28}\)

        \(=\frac{5}{2}\)

f) \(F=\left(9,75.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)

       \(=\left(\frac{39}{4}.21\frac{3}{7}+\frac{39}{4}.18\frac{4}{7}\right)\frac{15}{78}\)

       \(=\frac{39}{4}\left(21\frac{3}{7}+18\frac{4}{7}\right)\frac{15}{78}\)

       \(=\frac{39}{4}.40.\frac{15}{78}\)

       \(=390.\frac{15}{78}\)

       \(=78\)

Chúc bạn học tốt môn Toán!!!

Mình ko ghi đề đâu

\(A=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}=35-5\frac{7}{32}=29\frac{25}{32}\)

\(B=71\frac{38}{45}-43\frac{8}{45}+1\frac{17}{57}=\left(71\frac{38}{45}-43\frac{8}{45}\right)+1\frac{17}{57}=28\frac{2}{3}+1\frac{17}{57}=29\frac{55}{57}\)

\(C=-\frac{3}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}=-\frac{3}{7}.1+2\frac{3}{7}=-\frac{3}{7}+2\frac{3}{7}=2\)

a) \(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)

        \(=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

        \(=35-5\frac{7}{32}\) 

        \(=34\frac{32}{32}-5\frac{7}{32}\)       

        \(=29\frac{25}{32}\) 

b) \(B=71\frac{38}{45}-\left(43\frac{8}{45}-1\frac{17}{57}\right)\)

        \(=\left(71\frac{38}{45}-43\frac{8}{45}\right)+1\frac{17}{57}\)

        \(=28\frac{2}{3}+1\frac{17}{57}\)  

        \(=28\frac{38}{57}+1\frac{17}{57}\)

        \(=29\frac{55}{57}\)

c) \(C=\frac{-3}{7}.\frac{5}{9}+\frac{4}{9}.\frac{-3}{7}+2\frac{3}{7}\)

        \(=\frac{-3}{7}\left(\frac{5}{9}+\frac{4}{9}\right)+2\frac{3}{7}\)

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

        \(=2\)

d) \(D=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right)\frac{4}{5}\)

        \(=\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}.\frac{4}{5}\)

        \(=\frac{51}{8}:\frac{7}{12}.\frac{4}{5}\)

        \(=\frac{153}{14}.\frac{4}{5}\)

        \(=\frac{306}{35}\)

Ta có : \(A=8\frac{2}{7}-\left(3\frac{4}{9}+4\frac{2}{7}\right)\)

\(\Rightarrow A=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\)

\(\Rightarrow A=\frac{58}{7}-\frac{487}{63}=\frac{5}{9}\)

P/s:Câu B tương tự nhé

Tiếp B của @Phạm Tuấn Đạt

\(B=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)

\(\Rightarrow B=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)

\(B=\left(\frac{92}{9}-\frac{56}{9}\right)+\frac{13}{5}\)

\(B=\frac{36}{9}+\frac{13}{5}\)

\(B=4+\frac{13}{5}\)

\(B=\frac{20}{5}+\frac{13}{5}=\frac{33}{5}\)

a=58/7-(31/9+30/7)

a=58/7-(217/63+270/63)

a=58/7-487/63

a=5222/63-487/63

a=35/63

\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)

a)

\(\frac{{{4^3}{{.9}^7}}}{{{{27}^5}{{.8}^2}}} = \frac{{{{\left( {{2^2}} \right)}^3}.{{\left( {{3^2}} \right)}^7}}}{{{{\left( {{3^3}} \right)}^5}.{{\left( {{2^3}} \right)}^2}}} =\frac{2^{2.3}.3^{2.7}}{3^{3.5}.2^{2.3}}= \frac{{{2^6}{{.3}^{14}}}}{{{3^{15}}{{.2}^6}}} = \frac{1}{3}\)                  

b)

\(\frac{{{{\left( { - 2} \right)}^3}.{{\left( { - 2} \right)}^7}}}{{{{3.4}^6}}} =\frac{(-2)^{3+7}}{3.(2^2)^6}= \frac{{{{\left( { - 2} \right)}^{10}}}}{{3.{{\left( {{2^{2.6}}} \right)}}}} = \frac{{{2^{10}}}}{{{{3.2}^{12}}}} = \frac{1}{{{{3.2}^2}}} = \frac{1}{{12}}\)

c)

\(\begin{array}{l}\frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,09} \right)}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left[ {{{\left( {0,3} \right)}^2}} \right]}^3}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}} = \frac{{{{\left( {0,2} \right)}^5}.{{\left( {0,3} \right)}^6}}}{{{{\left( {0,2} \right)}^7}.{{\left( {0,3} \right)}^4}}}\\ = \frac{{{{\left( {0,3} \right)}^2}}}{{{{\left( {0,2} \right)}^2}}} = \frac{{0,9}}{{0,4}} = \frac{9}{4}\end{array}\)    

d)

Cách 1: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{8 + 16 + 32}}{{49}} = \frac{{56}}{{49}} = \frac{8}{7}\)

Cách 2: \(\frac{{{2^3} + {2^4} + {2^5}}}{{{7^2}}} = \frac{{2^3.(1+2+2^2)}}{{7^2}} = \frac{{2^3.7}}{{7^2}} = \frac{8}{7}\)