Gía trị biểu thức :\(\frac{7.\left(4^6.9^5+6^9.120\right)}{-8^4.3^{12}-6^{11}}\)
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XC
0
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)
=\(\frac{2.1.6}{1.3.5}=\frac{2.1.2}{1.1.5}=\frac{4}{5}\)
T
0
E
1
9 tháng 7 2019
\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{23}.3^{23}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{23}.3^{23}}=\frac{6}{2^{11}.3^{13}}=\frac{2.3}{2^{11}.3^{12}}=\frac{1}{2^{10}.3^{11}}=\frac{1}{6^{10}.3}\)
HX
0
M
2
27 tháng 9 2020
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{4}{5}\)
F
27 tháng 9 2020
Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}\)
\(=\frac{2\cdot6}{3\cdot5}=\frac{4}{5}\)
DT
1
29 tháng 11 2015
\(=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{\left(-6\right)^{11}-2^{12}\cdot3^{12}}=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{\left(-6\right)^{11}-6^{12}}=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{\left(-6\right)^{11}\left(1+6\right)}=\frac{6^{10}\cdot2^2\cdot2\cdot3}{\left(-6\right)^{11}\cdot7}=\frac{6^{11}\cdot4}{\left(-6\right)^{11}\cdot7}=\frac{-4}{7}\)
