Tính giá trị các biểu thức sau
4^6.9^5+6^9.120/8^4.3^12-6^11
K
Khách
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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)
=\(\frac{2.1.6}{1.3.5}=\frac{2.1.2}{1.1.5}=\frac{4}{5}\)
E
1
9 tháng 7 2019
\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{-2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{23}.3^{23}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{23}.3^{23}}=\frac{6}{2^{11}.3^{13}}=\frac{2.3}{2^{11}.3^{12}}=\frac{1}{2^{10}.3^{11}}=\frac{1}{6^{10}.3}\)
L
4
S
5 tháng 1 2023
\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)
V
6
R
7 tháng 2 2022
\(A=\dfrac{5}{7}.\dfrac{5}{11}+\dfrac{5}{7}.\dfrac{8}{11}-\dfrac{5}{7}.\dfrac{2}{11}\)
\(A=\dfrac{5}{7}.\left(\dfrac{5}{11}+\dfrac{8}{11}-\dfrac{2}{11}\right)\)
\(A=\dfrac{5}{7}.\dfrac{5+8-2}{11}\)
\(A=\dfrac{5}{7}.\dfrac{11}{11}\)
\(A=\dfrac{5}{7}.1=\dfrac{5}{7}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{35}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{63}\)
\(B=\dfrac{95}{72}\)
\(C=\dfrac{4^6.9^5+6^9.120}{8^4-3^{12}-6^{11}}\)
\(C=\dfrac{\left(2^2\right)^3.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(C=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(C=\dfrac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(C=\dfrac{2.6}{5.3}=\dfrac{12}{15}=\dfrac{4}{5}\)
XC
0
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.6}{3.5}\)
\(=\frac{4}{5}\)
Ta có: \(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}\)
\(=\frac{2\cdot6}{3\cdot5}=\frac{4}{5}\)