Tính hợp lí :
a ) 10 . \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}.6^{11}}\)
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TH
0
11 tháng 12 2018
\(\frac{10^3+2.5^3+5^3}{55}-\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)\(=\frac{2^3.5^3+2.5^3+5^3}{5.11}-\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{5^3\left(2^3+2+1\right)}{5.11}-\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}=\frac{5^2.11}{11}-\frac{2.6}{3.5}=25-\frac{4}{5}=\frac{121}{5}\)
MQ
3
MQ
3
13 tháng 2 2016
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)= \(\frac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.6.20}{\left(2^3\right)^4.3^{12}-6^{11}}\)= \(\frac{2^{12}.3^{10}+6^{10}.20}{2^{12}.3^{12}-6^{11}}\)
\(\frac{2^{10}.2^2.3^{10}+6^{10}.20}{6^{12}-6^{11}}\)= \(\frac{6^{10}\left(2^2+20\right)}{6^{10}\left(6^2-6\right)}\)= \(\frac{24}{30}\)= \(\frac{4}{5}\)
V
6
7 tháng 8 2016
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}.10=\frac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^{12}-2^{11}.3^{11}}.10=\frac{2^{10}.3^{10}\left(2^2+20\right)}{2^{10}.3^{10}\left(6^2-6\right)}.10=\frac{24.10}{30}=8\)
HL
1
CD
2
24 tháng 9 2017
Ta có : \(A=\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+2^9.3^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{-2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(-2.3-1\right)}=\frac{2\left(1+5\right)}{3\left(-6-1\right)}=\frac{2.6}{3.\left(-7\right)}=\frac{-12}{21}\)
Sửa đề : \(10.\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(10\cdot\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}+2\cdot9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=10\cdot\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)
\(=10\cdot\frac{2\cdot6}{3\cdot5}=8\)