So sánh A = \(\frac{3}{7}\left(\frac{3}{7}\right)^{19}\)và B = \(\left[\left(\frac{-3}{7}\right)^5\right]^4\)
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\(\begin{array}{l}a)\left[ {{{\left( {\dfrac{3}{7}} \right)}^4}.{{\left( {\dfrac{3}{7}} \right)}^5}} \right]:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{4 + 5}}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^9}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{9-7}}\\= {\left( {\dfrac{3}{7}} \right)^2}\\b)\left[ {{{\left( {\dfrac{7}{8}} \right)}^5}:{{\left( {\dfrac{7}{8}} \right)}^4}} \right].\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^{5 - 4}}.\left( {\dfrac{7}{8}} \right)\\ = \left( {\dfrac{7}{8}} \right).\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^2}\\c)\left[ {{{\left( {0,6} \right)}^3}.{{\left( {0,6} \right)}^8}} \right]:\left[ {{{\left( {0,6} \right)}^7}.{{\left( {0,6} \right)}^2}} \right]\\ = {\left( {0,6} \right)^{3 + 8}}:{\left( {0,6} \right)^{7 + 2}}\\ = {\left( {0,6} \right)^{11}}:{\left( {0,6} \right)^9}\\ = {\left( {0,6} \right)^{11-9}}\\={\left( {0,6} \right)^2}.\end{array}\)
Ta có
\(A=\frac{\left(3\frac{2}{5}+\frac{1}{5}\right):2\frac{1}{2}}{\left(5\frac{3}{7}-2\frac{1}{4}\right):4\frac{43}{56}}\) \(B=\frac{1,2:\left(1\frac{1}{5}-1\frac{1}{4}\right)}{0,32+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\left(\frac{17}{5}+\frac{1}{5}\right):\frac{5}{2}}{\left(\frac{38}{7}-\frac{9}{4}\right):\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(\frac{6}{5}-\frac{5}{4}\right)}{\frac{8}{25}+\frac{2}{25}}\)
\(\Leftrightarrow A=\frac{\frac{18}{5}:\frac{5}{2}}{\frac{89}{28}:\frac{276}{56}}\) \(\Leftrightarrow B=\frac{\frac{6}{5}:\left(-\frac{1}{20}\right)}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{\frac{36}{25}}{\frac{89}{138}}\) \(\Leftrightarrow B=\frac{\frac{5}{4}}{\frac{2}{5}}\)
\(\Leftrightarrow A=\frac{4968}{2225}\) \(\Leftrightarrow B=\frac{25}{8}\)
\(\Leftrightarrow A=\frac{39744}{17800}\) \(\Leftrightarrow B=\frac{55625}{17800}\)
Ta có: 39744<55625
\(\Rightarrow A< B\)
Vậy A<B
\(a)\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{-4}{3}}\)\(=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{3}{10}.\frac{-4}{3}}=\frac{5}{24}\)
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