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\(P=\frac{3^9\cdot3^{20}\cdot2^8}{3^{24}\cdot243\cdot2^6}\)
\(P=\frac{3^{29}\cdot2^8}{3^{29}\cdot2^6}\)
\(P=2^2=4\)
\(Q=\frac{2^{15}\cdot5^3\cdot2^6\cdot3^4}{8\cdot2^{18}\cdot81\cdot5}\)
\(Q=\frac{2^{21}\cdot5^3\cdot3^4}{2^{21}\cdot3^4\cdot5}\)
\(Q=5^2=25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a 25 phần 41
b 9
c 1 phần 12
d 12
e 14 phần 15
f 24 phần 7
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Tính
a)
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}\\ =\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}....\frac{99.101}{100}\\ \)
\(=\left(\frac{1.2.3...99}{2.3...100}\right).\left(\frac{3.4.5...101}{2.3.4...100}\right)\\ =\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
b)
\(\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{n^2}\\ < \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(n-1\right)n}\\ \)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{n-1}-\frac{1}{n}\\ =1-\frac{1}{n}< 1\)
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\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
Bài làm của mk hơi tắt nên bạn tự suy luận nhé
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)=\(\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)=\(\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1+5\right)}\)=\(\frac{-13122}{6561.6}\)=\(-\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`(3^9. 3^20. 2^8)/(3^24. 243. 2^6)`
`=(3^29. 2^8)/(3^24. 3^5. 2^6)`
`=(3^29. 2^8)/(3^29. 2^6)`
`=2^2=4`
\(\frac{1+2^4+2^8+.....+2^{20}}{2^4+2^8+.......+2^{24}}=\frac{1+2^4+2^8+....+2^{20}}{2^4\cdot\left(1+2^4+2^8+...+2^{20}\right)}\)
\(=\frac{1}{2^4}=\frac{1}{16}\)