\(\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^2-6^{11}}\)
. là nhân nhé . Đề bài thính hợp lý
nhanh gọn lẹ cho mik nge
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Những câu hỏi liên quan
NP
3
N
18 tháng 10 2017
\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.2^3.3.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}}{2^{11}.3^{11}}.\frac{1+5}{2.3-1}\)
\(=\frac{2^{12}.3^{10}}{2^{11}.3^{11}}.\frac{6}{5}\)
\(=\frac{2}{3}.\frac{6}{5}\)
\(=\frac{4}{5}\)
NH
2
19 tháng 12 2021
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}=\dfrac{2^{13}\cdot3^{11}}{2^{11}\cdot3^{11}\cdot7}=\dfrac{4}{7}\)
KT
0
NV
2
MT
1
2 tháng 2 2020
\(A=\frac{4^6\cdot9^5+6^9\cdot120}{8^4\cdot3^{12}-6^{11}}=\frac{\left(2^2\right)^6\cdot\left(3^2\right)^5+\left(2\cdot3\right)^9\cdot2^3\cdot3\cdot5}{\left(2^3\right)^4\cdot3^{12}-\left(2\cdot3\right)^{11}}=\frac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\frac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=\frac{2\left(1+5\right)}{3\left(6-1\right)}=\frac{2\cdot6}{3\cdot5}=\frac{2\cdot2}{5}=\frac{4}{5}\)
\(\approx GOOD\)\(LUCK\approx\)
NP
0
M
16 tháng 5 2017
Ta có :
\(M=\frac{9^4.27^5.3^6.3^4}{3^8.81^4.23^4.8^2}\)
\(M=\frac{\left(3^2\right)^4.\left(3^3\right)^5.3^{10}}{3^8.\left(3^4\right)^4.23^4.8^2}\)
\(M=\frac{3^8.3^{15}.3^{10}}{3^8.3^{16}.23^4.8^2}\)
\(M=\frac{3^{33}}{3^{24}.23^4.8^2}\)
\(M=\frac{3^9}{23^4.8^2}\)
21 tháng 4 2019
Bài 1
a) \(P=\frac{6n+5}{2n-4}=\frac{6n-12+7}{2n-4}=3+\frac{7}{2n-4}\)
Để P là phân số thì \(\hept{\begin{cases}2n-4\ne7\\2n-4\ne1\end{cases}}\Leftrightarrow\hept{\begin{cases}n\ne\frac{11}{2}\\n\ne\frac{5}{2}\end{cases}}\)
Vậy...
b) \(P=\frac{6n+5}{2n-4}=3+\frac{7}{2n-4}\)
Để \(P\in Z\)thì \(\orbr{\begin{cases}2n-4=7\\2n-4=1\end{cases}\Leftrightarrow\orbr{\begin{cases}n=\frac{11}{2}\notin Z\\n=\frac{5}{2}\notin Z\end{cases}}}\)
Vậy không có giá trị n nào thuộc Z để P thuộc Z.
c) \(\left|2n-3\right|=\frac{5}{3}\)
Trường hợp: \(2n-3=\frac{5}{3}\Rightarrow n=\frac{7}{3}\)
\(P=\frac{6.\frac{7}{3}+5}{2.\frac{7}{3}-4}=\frac{19}{\frac{2}{3}}=\frac{57}{2}\)
Trường hợp: \(2n-3=-\frac{5}{3}\Rightarrow n=\frac{2}{3}\)
\(P=\frac{6.\frac{2}{3}+5}{2.\frac{2}{3}-4}=\frac{9}{\frac{-8}{3}}=\frac{27}{-8}\)
Bài 2
\(N=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^{10}.4.5}{\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)
\(=\frac{2^{12}.3^{10}+5.2^{12}.3^{10}}{2^{12}.3^{12}-6^{11}}=\frac{6.2^{12}.3^{10}}{6^{12}-6^{11}}\)
\(=\frac{2.3.2^{12}.3^{10}}{6.6^{11}-6^{11}}=\frac{2^{13}.3^{11}}{5.\left(2.3\right)^{11}}=\frac{2^{13}.3^{11}}{5.2^{11}.3^{11}}=\frac{4}{5}\)
\(\frac{4^6.9^5+6^9.120}{8^4.3^2-6^{11}}\)
\(=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^2-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^2.\left(2-3^9\right)}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^2.\left(2-3^9\right)}\)
\(=\frac{2.3^8.6}{2-3^9}\)
\(=\frac{2.2.3.3^8}{2-3^9}\)
\(=\frac{2^2.3^9}{2-3^9}\)