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\(\frac{15.27^7.3^8-9^{15}}{4^2.81^8}=\frac{3.5.\left(3^3\right)^7.3^8-\left(3^2\right)^{15}}{\left(2^2\right)^2.\left(3^4\right)^8}\)
\(=\frac{3.5.3^{21}.3^8-3^{30}}{2^4.3^{24}}=\frac{3^{32}.5-3^{30}}{2^4.3^{24}}\)
\(=\frac{3^{30}\left(3^2.5-1\right)}{2^4.3^{24}}=\frac{3^6.44}{2^4}=\frac{3^6.2^2.11}{2^4}=\frac{3^6.11}{2^2}\)
Ta có :\(\frac{15.27^7.3^8-9^{15}}{4^2.81^8}=\frac{3.5.\left(3^3\right)^7.3^8-\left(3^2\right)^{15}}{4^2.\left(3^4\right)^8}=\frac{3.5.3^{21}.3^8-3^{30}}{4^2.3^{32}}=\frac{3^{30}.5-3^{30}}{4^2.3^{32}}=\frac{3^{30}.4}{4^2.3^{32}}=\frac{1}{4.3^2}=\frac{1}{36}\)
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) \(=\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}.4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}\)
\(=\frac{13}{56}\)
b) \(=\frac{3}{2}+\frac{1}{2}.\frac{7}{18}:\frac{7}{12}\)
\(=\frac{3}{2}+\frac{1}{3}\)
\(=\frac{11}{6}\)
13/50+9/100+41/100+12/50
=(13/50+12/50)+(9/100+41/100)
=1/2+1/2
=1
11) Ta có:
\(\frac{120-0,5.40.5.0,2.20.0,25-20}{1+5+9+...+33+37}\)
\(=\frac{120-\left(0,5.40\right).\left(5.0,2\right).\left(20.0,25\right)-20}{1+5+9+...+33+37}\)
\(=\frac{120-20.1.5-20}{1+5+9+...+33+37}\)
\(=\frac{120-100-20}{1+5+9+...+33+37}\)
\(=\frac{0}{1+5+9+...+33+37}=0\)
Ta có : \(\frac{9^9.32^9}{64^7.27^6}=\frac{\left(3^2\right)^9.\left(2^5\right)^9}{\left(2^7\right)^7.\left(3^3\right)^6}=\frac{3^{18}.2^{45}}{2^{49}.3^{18}}=\frac{1}{2^4}=\frac{1}{16}\)