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10 tháng 7 2016
\(4^6.9^5+6^9.120:8^4.3^{12}+6^{11}=\frac{4^6.9^5+6^9.120}{8^4.3^{12}+6^{11}}=\frac{ \left(2^2\right)^6.\left(3^2\right)^5+6^9.6.20}{2^3.3^{12}+6^9.6.6}=\frac{2^{12}.3^{10}+20}{2^3.3^{12}+6}=\frac{2^9+20}{3^2+6}\)
\(\frac{2^9+20}{3^2+6}=\frac{512+20}{9+6}=\frac{532}{15}\)
1 tháng 7 2018
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}+6^{11}}\) \(=\) \(\frac{\left(2^2\right)^6.\left(3^2\right)^5+2^9.3^9.2^3.3.5}{\left(2^3\right)^4.3^{12}+2^{11}.3^{11}}\) \(=\) \(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\) \(\frac{2^{11}.3^{10}.\left(2+2.5\right)}{2^{11}.3^{10}.\left(2.3^2+3\right)}\) \(=\) \(\frac{2+10}{2.9+3}\) \(=\) \(\frac{12}{21}\) \(=\) \(\frac{4}{7}\)
V
6
KN
8
BY
4 tháng 7 2016
\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{6^{12}-6^{11}}\)
\(=\frac{2^{12}3^{10}\left(1+5\right)}{6^{11}\left(6-1\right)}\)
\(=\frac{2^{10}\cdot3^{10}\cdot5\cdot2^2}{6^{10}\cdot6\cdot5}\)
\(=\frac{6^{10}\cdot20}{6^{10}\cdot30}\)
\(=\frac{2}{3}\)
NM
4 tháng 7 2016
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\) (Sau đó phân tách ra)
=\(\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
=\(\frac{2^{12}.3^{10}+2^9.3^9.2^33.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
=\(\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\) (Gộp và giải như biểu thức thường)
=\(\frac{2^{12}.3^{10}.\left(1+1.5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
=\(\frac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\) (Rút gọn giữa tử và mẫu)
=\(\frac{2.1.6}{1.3.5}=\frac{2.1.2}{1.1.5}=\frac{4}{5}\)
LC
4
NH
13 tháng 8 2020
\(10\cdot\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}+2\cdot9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=10\cdot\frac{2\cdot6}{3\cdot5}=8\)
7 tháng 8 2016
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}.10=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}-2^{11}.3^{11}}.10=\frac{2^{10}.3^{10}\left(2^2+20\right)}{2^{10}.3^{10}\left(6^2-6\right)}.10=\frac{24.10}{30}=8\)
HN
1
HH
24 tháng 6 2018
Giải:
\(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=10.\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10.\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10.\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=10.\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\)
\(=10.\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.5}\)
\(=10.\dfrac{2^2}{5}\)
\(=2^3=8\)
Vậy ...
HL
1