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

khó thấy quá bạn ơi

a) \(0,2 + 2,5:\frac{7}{2} = \frac{2}{{10}} + \frac{25}{10}:\frac{7}{2} = \frac{1}{5} + \frac{25}{10}.\frac{2}{7} \\= \frac{1}{5} + \frac{5}{7} = \frac{7}{{35}} + \frac{{25}}{{35}} = \frac{{32}}{{35}}\)

b)

\(\begin{array}{l}9.{\left( {\frac{{ - 1}}{3}} \right)^2} - {\left( { - 0,1} \right)^3}:\frac{2}{{15}}\\ = 9.\frac{1}{9} - {\left( {\frac{{ - 1}}{{10}}} \right)^3}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}:\frac{2}{{15}}\\ = 1 - \frac{{ - 1}}{{1000}}.\frac{{15}}{2}\\ = 1 + \frac{3}{{400}}\\=\frac{400}{400}+\frac{3}{400}\\ = \frac{{403}}{{400}}\end{array}\)

a)

 \(\begin{array}{l}\left( {\frac{{ - 2}}{{ - 5}}:\frac{3}{{ - 4}}} \right).\frac{4}{5} = \left( {\frac{2}{5}.\frac{{ - 4}}{3}} \right).\frac{4}{5}\\ = \frac{{ - 8}}{{15}}.\frac{4}{5} = \frac{{ - 32}}{{75}}\end{array}\)

b)

\(\begin{array}{l}\frac{{ - 3}}{{ - 4}}:\left( {\frac{7}{{ - 5}}.\frac{{ - 3}}{2}} \right) = \frac{3}{4}:\frac{{ - 21}}{{ - 10}}\\ = \frac{3}{4}.\frac{{10}}{{21}} = \frac{{30}}{{84}} = \frac{5}{14}\end{array}\)

c)

 \(\begin{array}{l}\frac{{ - 1}}{9}.\frac{{ - 3}}{5} + \frac{5}{{ - 6}}.\frac{{ - 3}}{5} + \frac{5}{2}.\frac{{ - 3}}{5}.\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 1}}{9} + \frac{5}{{ - 6}} + \frac{5}{2}} \right)\\ = \frac{{ - 3}}{5}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 15}}{{18}} + \frac{{45}}{{18}}} \right)\\ = \frac{{ - 3}}{5}.\frac{{28}}{{18}}\\ = \frac{{ - 3}}{5}.\frac{{14}}{9}\\ = \frac{{ - 14}}{{15}}\end{array}\)