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A=(\(\frac{3}{2}\)-\(\frac{2}{5}\)+\(\frac{1}{10}\)) : (\(\frac{3}{2}\)-\(\frac{2}{3}\)+\(\frac{1}{12}\))
A=(\(\frac{15}{10}\)-\(\frac{4}{10}\)+\(\frac{1}{10}\)):( \(\frac{18}{12}\)-\(\frac{8}{12}\)+ \(\frac{1}{12}\))
A=\(\frac{12}{10}\):\(\frac{11}{12}\)=\(\frac{12}{10}\).\(\frac{12}{11}\)=\(\frac{72}{55}\)
\(A=\frac{\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right)}{\left(\frac{3}{2}-\frac{2}{3}+\frac{1}{12}\right)}\)
\(A=\frac{\left(\frac{15}{10}-\frac{4}{10}+\frac{1}{10}\right)}{\left(\frac{18}{12}-\frac{8}{12}+\frac{1}{12}\right)}\)
\(A=\frac{\frac{6}{5}}{\frac{11}{12}}=\frac{6}{5}:\frac{11}{12}=\frac{6}{5}\times\frac{12}{11}\)
\(A=\frac{72}{55}\)
BÀI 1
\(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.\left(3^2\right)^3}{\left(2.3\right)^5.\left(2^3\right)^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}.\)
bài 2
a) \(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{3}{12}=\frac{1}{4}\)
b) \(\frac{9^9.27^4}{3^8.81^5}=\frac{\left(3^2\right)^9.\left(3^3\right)^4}{3^8.\left(3^4\right)^5}=\frac{3^{18}.3^{12}}{3^8.3^{20}}=\frac{3^{30}}{3^{28}}=3^2=9\)
Study well
Bài 1: \(\frac{2^7.9^3}{6^5.8^2}=\frac{2^7.3^6}{2^5.3^5.2^6}=\frac{3}{2^4}=\frac{3}{16}\)
Bài 2:
a)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{12}=\frac{6}{12}-\frac{4}{12}+\frac{1}{12}=\frac{6-4+1}{12}=\frac{1}{4}\)
b)\(\frac{9^9.27^4}{3^8.81^5}=\frac{9^9.3^{12}}{3^8.9^{10}}=\frac{3^4}{9}=\frac{3^4}{3^2}=3^2=9\)
C = \(\frac{2}{3}\sqrt{144}-\left(-\frac{3}{4}\right)\div\sqrt{\frac{225}{144}}\)
C = \(\frac{2}{3}.12+\frac{3}{4}\div\frac{5}{4}\)
C = \(8+\frac{3}{5}\)
C = \(8\frac{3}{5}\)
D = \(\frac{4^6.25^5-2^{12}.25^4}{2^{12}.5^8-10^8.64}\)
D = \(\frac{\left(2^2\right)^6.\left(5^2\right)^5-2^{12}.\left(5^2\right)^4}{2^{12}.5^8-\left(2.5\right)^8.2^6}\)
D = \(\frac{2^{12}.5^{10}-2^{12}.5^8}{2^{12}.5^8-2^8.5^8.2^6}\)
D = \(\frac{2^{12}.5^8.\left(25-1\right)}{2^{12}.5^8.\left(1-2^2\right)}\)
D = \(\frac{24}{-3}\)
D = \(-8\)
\(C=\frac{2}{3}\sqrt{144}-\left(\frac{-3}{4}\right):\sqrt{\frac{225}{144}}\)
\(=\frac{2}{3}\cdot12+\frac{3}{4}:\frac{5}{4}\)
\(=8+\frac{3}{4}\cdot\frac{4}{5}\)
\(=8+\frac{3}{5}\)
\(=\frac{40}{5}+\frac{3}{4}=\frac{43}{5}\)
\(D=\frac{4^6\cdot25^5-2^{12}\cdot25^4}{2^{12}\cdot5^8-10^8\cdot64}=\frac{\left(2^2\right)^6\cdot\left(5^2\right)^5-2^{12}\cdot\left(5^2\right)^4}{2^{12}\cdot5^8-\left(2\cdot5\right)^8\cdot2^6}\)
\(=\frac{2^{12}\cdot5^{10}-2^{12}\cdot5^8}{2^{12}\cdot5^8-2^{14}\cdot5^8}=\frac{5^8\left(2^{12}\cdot5^2-2^{12}\right)}{5^8\left(2^{12}-2^{14}\right)}\)
\(=\frac{2^{12}\cdot5^2-2^{12}}{2^{12}-2^{14}}=\frac{2^{12}\left(5^2-1\right)}{2^{12}\left(1-2^2\right)}=\frac{24}{-3}=-8\)
A=\(\left(\frac{3}{2}-\frac{2}{5}+\frac{1}{10}\right).\left(\frac{2}{3}-\frac{3}{2}+12\right)\)
A=\(\frac{6}{5}\).\(\frac{67}{6}\)=\(\frac{67}{5}\)
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