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\(2^5\).\(9^5\).\(2^8\).\(9^8\)

=(\(2^5\).\(2^8\)).(\(9^5\).\(9^8\))

=\(^{2^{13}}\).\(9^{13}\)

=\(^{2.9^{13}}\)

=\(18^{13}\)

21 tháng 10 2018

ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac

28 tháng 1 2019

\(A=\frac{2^{12}.3^4-4^5.9^2}{\left(2^2.3\right)^6+8^4.3^5}\)

\(A=\frac{2^{12}.3^4-2^{10}.3^4}{2^{12}.3^6+2^{12}.3^5}\)

\(A=\frac{2^{10}.3^4\left(2^2-1\right)}{2^{10}.3^4\left(2^2.3^2+2^2.3\right)}\)

\(A=\frac{2^2-1}{2^2.3^2+2^2.3}\)

\(A=\frac{4-1}{36+12}\)

\(A=\frac{3}{48}=\frac{1}{16}\)

19 tháng 3 2021

\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}:\sqrt{\dfrac{25}{9}}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}:\dfrac{5}{3}\)

\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\dfrac{5}{3}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}.\dfrac{5}{3}=\dfrac{1-3}{1-5}.\dfrac{5}{3}=\dfrac{1}{2}.\dfrac{5}{3}=\dfrac{5}{6}\)

19 tháng 3 2021

\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\div\sqrt{\dfrac{25}{9}}\)

\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\div\dfrac{5}{4}\)

=\(\dfrac{2^{10}\cdot3^8\left(1-2\cdot3\right)}{2^{10}\cdot3^8\left(1+5\right)}\div\dfrac{5}{4}\)

=\(\dfrac{1-6}{1+5}\cdot\dfrac{4}{5}\)

=\(-\dfrac{5}{6}\cdot\dfrac{4}{5}\)

=\(-\dfrac{2}{3}\)

\(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)

\(\dfrac{4^{20}\cdot3^{35}}{2^{37}\cdot27^{12}}=\dfrac{\left(2^2\right)^{20}\cdot3^{35}}{2^{37}\cdot\left(3^3\right)^{12}}=\dfrac{2^{40}\cdot3^{35}}{2^{37}\cdot3^{36}}=\dfrac{2^3}{3}\)

\(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^5\cdot5^5\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{100}\)

\(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=3^2\)

12 tháng 7 2016

\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)

\(M=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)

\(M=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)

\(M=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)

\(M=2^{10}\)

\(M=1024\)

22 tháng 7 2016

\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\times\left(2^{20}+1\right)}{2^{30}\times\left(2^{20}+1\right)}=2^{10}=1024\)

Chúc bạn học tốt ^^

1 tháng 12 2016

- Có thể cho tôi một số bài tập dạng này không?

30 tháng 10 2020

1. \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25=5^2\Leftrightarrow x=5\)

2. \(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=\frac{2^{40}}{2^{30}}=2^{10}\)

1)\(x^{10}=25x^8\)

\(\Rightarrow x^{10}:x^8=25\)

\(\Rightarrow x^2=5^2\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)

2)\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)