Tính
\(^{2^5.9^5.}^{2^8.}^{9^8}\)
\(\frac{8^{20}+4^{20}}{4^{25}+64^5}\)
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21 tháng 10 2018
ban nao co chuyen shin ko cho minh muon minh giai cho 10 bai nhu the i love pac pac
LN
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
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\div\sqrt{\dfrac{25}{9}}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\div\dfrac{5}{4}\)
=\(\dfrac{2^{10}\cdot3^8\left(1-2\cdot3\right)}{2^{10}\cdot3^8\left(1+5\right)}\div\dfrac{5}{4}\)
=\(\dfrac{1-6}{1+5}\cdot\dfrac{4}{5}\)
=\(-\dfrac{5}{6}\cdot\dfrac{4}{5}\)
=\(-\dfrac{2}{3}\)
1 tháng 10 2023
\(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)
\(\dfrac{4^{20}\cdot3^{35}}{2^{37}\cdot27^{12}}=\dfrac{\left(2^2\right)^{20}\cdot3^{35}}{2^{37}\cdot\left(3^3\right)^{12}}=\dfrac{2^{40}\cdot3^{35}}{2^{37}\cdot3^{36}}=\dfrac{2^3}{3}\)
\(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^5\cdot5^5\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{100}\)
\(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=3^2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
27 tháng 7
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (\(\dfrac{2}{2}\) + \(\dfrac{3}{3}\) + \(\dfrac{4}{4}\) + \(\dfrac{5}{5}\)+ \(\dfrac{6}{6}+\dfrac{7}{7}+\dfrac{8}{8}\) + \(\dfrac{10}{10}\))
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x (1 + 1 +1 + 1+ 1+ 1+ 1 +1)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) x 1 x 8
= \(\dfrac{1}{2}\) + \(\)\(\dfrac{1}{2}\) x 8
= \(\dfrac{1}{2}\) + 4
= \(\dfrac{9}{2}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
a; \(\dfrac{1}{4}\) + \(\dfrac{2}{5}\) + \(\dfrac{6}{8}\) + \(\dfrac{9}{15}\) + \(\dfrac{8}{1}\)
= (\(\dfrac{1}{4}\) + \(\dfrac{6}{8}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{9}{15}\)) + 8
= (\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\)) + (\(\dfrac{2}{5}\) + \(\dfrac{3}{5}\)) + 8
= 1 + 1 + 8
= 2 + 8
= 10
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
21 tháng 9
b; \(\dfrac{1}{2}\) + \(\dfrac{2}{4}\) + \(\dfrac{3}{6}\) + \(\dfrac{4}{8}\) + \(\dfrac{5}{10}\) + \(\dfrac{6}{12}\) + \(\dfrac{7}{14}\) + \(\dfrac{8}{16}\) + \(\dfrac{9}{18}\) + \(\dfrac{10}{20}\)
= \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)
= \(\dfrac{1}{2}\) x 10
= 5
12 tháng 7 2016
\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)
\(M=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)
\(M=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)
\(M=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)
\(M=2^{10}\)
\(M=1024\)
PA
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 ^^
N
2
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. \(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=\frac{2^{40}}{2^{30}}=2^{10}\)
30 tháng 10 2020
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)\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)
\(2^5\).\(9^5\).\(2^8\).\(9^8\)
=(\(2^5\).\(2^8\)).(\(9^5\).\(9^8\))
=\(^{2^{13}}\).\(9^{13}\)
=\(^{2.9^{13}}\)
=\(18^{13}\)