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a: \(A=\left(0+1\right):\left(\dfrac{2}{3}+\dfrac{7}{6}-\dfrac{1}{6}\right)=1:\dfrac{5}{3}=\dfrac{3}{5}\)
b: \(B=\left[0.8\cdot15\right]\cdot\left[1.25\cdot\dfrac{19}{3}\right]+31.64=15\cdot\dfrac{95}{12}+31.64=150.39\)
\(a,A=\left[\frac{4}{11}.\left(\frac{1}{25}\right)^0+\frac{7}{22}.2\right]^{2010}-\left(\frac{1}{2^2}:\frac{8^2}{4^4}\right)^{2009}\)
\(A=\left(\frac{4}{11}.1+\frac{7}{11}\right)^{2010}-\left(\frac{1}{2^2}.2^2\right)^{2009}\)
\(A=1-1=0\)
\(b,B=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)
\(B=\frac{0,8:1}{\frac{3}{5}}+\frac{\left(1\right):\frac{4}{7}}{\left(\frac{59}{9}-\frac{13}{4}\right).36}\)
\(B=0,8.\frac{5}{3}+\frac{\frac{7}{4}}{\frac{119}{36}.36}\)
\(B=\frac{4}{3}+\frac{7}{4}.\frac{1}{119}\)
\(B=\frac{4}{3}+\frac{1}{68}=\frac{275}{204}\)
Ta có: \(\frac{0,8:\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\cdot2\frac{2}{17}}+\frac{\left(1,2\cdot0,5\right)}{\frac{4}{5}}\)
\(=\frac{\frac{4}{5}:\left(\frac{4}{5}\cdot\frac{5}{4}\right)}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right)\cdot\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right)\cdot\frac{36}{17}}+\frac{6}{5}\cdot\frac{1}{2}\cdot\frac{5}{4}\)
\(=\frac{\frac{4}{5}}{\frac{3}{5}}+\frac{\frac{7}{4}}{\frac{119}{36}\cdot\frac{36}{17}}+\frac{3}{4}\)
\(=\frac{4}{5}\cdot\frac{5}{3}+\frac{7}{4}\cdot\frac{1}{7}+\frac{3}{4}=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}=\frac{7}{3}\)
\(A=\left[0,8.7\left(0,8\right)^2\right].\left(1,25.7-\frac{4}{5}.1,25\right)+31,64\)
\(=0,8.\left(7+0,8\right).1,25.\left(7-0,8\right)+31,64\)
\(=0,8.7,8.1,25.6,2+31,64\)
\(=6,24.7,75+31,64\)
\(=48,36+31,64=80\)
\(B=\frac{\left(1,09-0,29\right).\frac{5}{4}}{\left(18,9-16,65\right).\frac{8}{9}}=\frac{0,8.1,25}{2,25.\frac{8}{9}}=\frac{1}{2}\)
\(A:B=80:\frac{1}{2}=160\)
Vậy A gấp 160 lần B.
\(\begin{array}{l}a)\left( {\frac{2}{3} + \frac{1}{6}} \right):\frac{5}{4} + \left( {\frac{1}{4} + \frac{3}{8}} \right):\frac{5}{2}\\ = \left( {\frac{4}{6} + \frac{1}{6}} \right).\frac{4}{5} + \left( {\frac{2}{8} + \frac{3}{8}} \right).\frac{2}{5}\\ = \frac{5}{6}.\frac{4}{5} + \frac{5}{8}.\frac{2}{5}\\ = \frac{2}{3} + \frac{1}{4}\\ = \frac{8}{{12}} + \frac{3}{{12}}\\ = \frac{{11}}{{12}}\\b)\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{2}{7}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{7}{4}.\left( {\frac{1}{{14}} - \frac{4}{{14}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{7}{4}.\frac{{ - 3}}{{14}}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{{ - 3}}{8}\\ = \frac{{ - 110}}{{27}} + \frac{{ - 3}}{8}\\ = \frac{{ - 880}}{{216}} + \frac{{ - 81}}{{216}}\\ = \frac{{ - 961}}{{216}}\end{array}\)
a: \(A=\dfrac{1}{9}:\dfrac{1}{9}:\left(\dfrac{10+7}{15}:\dfrac{12-5}{30}\right)\)
\(=1:\left(\dfrac{17}{15}\cdot\dfrac{30}{7}\right)=1:\dfrac{34}{7}=\dfrac{7}{34}\)
b: \(=\left(5.6+0.64\right)\cdot1.25\cdot\dfrac{19}{3}+31.64\)
\(=\dfrac{39}{5}\cdot\dfrac{19}{3}+\dfrac{791}{25}=\dfrac{2026}{25}\)