\(\left(5\frac{4}{9}+1\frac{1}{3}\right)-3\frac{4}{9}\)
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\(A = {1\over2}-{3\over4}+{5\over6}-{7\over12}={6\over12}-{9\over12}+{10\over12}-{7\over12}\)\(={0\over12}=0\)
g) \(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}\)
\(=\left(6\frac{4}{5}-3\frac{4}{5}\right)-1\frac{2}{3}\)
\(=\left(6+\frac{4}{5}-3-\frac{4}{5}\right)-1\frac{2}{3}\)
\(=3-\frac{5}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)
h) \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=7\frac{5}{9}-2\frac{3}{4}-3\frac{5}{9}\)
\(=\left(7\frac{5}{9}-3\frac{5}{9}\right)-2\frac{3}{4}\)
\(=\left(7+\frac{5}{9}-3-\frac{5}{9}\right)-2\frac{3}{4}\)
\(=4-\frac{11}{4}=\frac{16}{4}-\frac{11}{4}=\frac{5}{4}\)
i) \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6\frac{5}{7}-1\frac{3}{4}-2\frac{5}{7}\)
\(=\left(6\frac{5}{7}-2\frac{5}{7}\right)-1\frac{3}{4}\)
\(=\left(6+\frac{5}{7}-2-\frac{5}{7}\right)-\frac{7}{4}\)
\(=4-\frac{7}{4}=\frac{16}{4}-\frac{7}{4}=\frac{9}{4}\)
k) \(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(=7\frac{5}{11}-2\frac{3}{7}-3\frac{5}{11}\)
\(=\left(7\frac{5}{11}-3\frac{5}{11}\right)-2\frac{3}{7}\)
\(=4-\frac{17}{7}=\frac{28}{7}-\frac{17}{7}=\frac{11}{7}\)
g) Ta có: \(6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\)
\(=\frac{34}{5}-\left(\frac{5}{3}+\frac{19}{5}\right)\)
\(=\frac{34}{5}-\frac{5}{3}-\frac{19}{5}\)
\(=3-\frac{5}{3}=\frac{9}{3}-\frac{5}{3}=\frac{4}{3}\)
h) Ta có: \(7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(=\frac{68}{9}-\frac{11}{4}-\frac{32}{9}\)
\(=4-\frac{11}{4}=\frac{16}{4}-\frac{11}{4}=\frac{5}{4}\)
i) Ta có: \(6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\)
\(=6+\frac{5}{7}-1-\frac{3}{4}-2-\frac{5}{7}\)
\(=3-\frac{3}{4}=\frac{9}{4}\)
k) Ta có: \(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(=7+\frac{5}{11}-2-\frac{3}{7}-3-\frac{5}{11}\)
\(=2-\frac{3}{7}=\frac{11}{7}\)
a, \(\frac{\left(\frac{1}{9}\right)^6\cdot\left(\frac{3}{8}\right)^7}{\left(\frac{1}{3}\right)^{13}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{\left(\frac{1}{\left(3^2\right)^6}\right)\cdot\left(\frac{1}{2}\cdot\frac{1}{2}\cdot\frac{1}{2}\cdot3\right)^7}{\left(\frac{1}{3}\right)^{13}.\left(\frac{1}{2}\right)^{22}.3^6}=\frac{\frac{1}{3^{12}}\cdot\left(\frac{1}{2}\right)^{21}\cdot3^7}{\frac{1}{3^{13}}\cdot\left(\frac{1}{2}\right)^{22}.3^6}\)
\(=\frac{3}{\frac{1}{3}\cdot\frac{1}{2}}=3\div\frac{1}{6}=3.6=18\)
b, Làm tương tự nha bn
\(\left(5\frac{4}{9}+1\frac{1}{3}\right)-3\frac{4}{9}\)
\(=5\frac{4}{9}-3\frac{4}{9}+1\frac{1}{3}\)
\(=2+\frac{4}{3}\)
\(=2\frac{4}{3}\)
\(\left(5\frac{4}{9}+1\frac{1}{3}\right)-3\frac{4}{9}\)
\(=2+1\frac{1}{3}\)
\(=\frac{10}{3}\)
P/s tham khảo nha