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13 tháng 7 2017
f) \(\frac{25^2.20^4}{5^{10}.4^5}=\frac{\left(5^2\right)^5.\left(4.5\right)^4}{5^{10}.4^5}=\frac{5^{10}.5^4.4^4}{5^{10}.4^5}=\frac{5^{14}.4^4}{5^{10}.4^5}=\frac{5^4}{4}\)
i) \(\frac{9^{15}.81^4}{27^8.3^{20}}=\frac{\left(3^2\right)^{15}.\left(3^4\right)^4}{\left(3^3\right)^8.3^{20}}=\frac{3^{30}.3^{16}}{3^{24}.3^{20}}=\frac{3^{46}}{3^{44}}=3^2=9\)
13 tháng 7 2017
f) Ta có: \(\frac{25^2.20^4}{5^{10}.4^5}\)= \(\frac{\left(5^2\right)^2.\left(4.5\right)^4}{5^{10}.4^5}\)= \(\frac{5^4.4^4.5^4}{5^{10}.4^5}\)= \(\frac{5^8.4^4}{5^{10}.4^5}\)= \(\frac{1}{5^2.4}\)=\(\frac{1}{100}\).
i) Ta có: \(\frac{9^{15}.81^4}{27^8.3^{20}}\)= \(\frac{\left(3^2\right)^{15}.\left(3^4\right)^4}{\left(3^3\right)^8.3^{20}}\)= \(\frac{3^{30}.3^8}{3^{24}.3^{20}}\)= \(\frac{3^{38}}{3^{44}}\)=\(\frac{1}{3^6}\)= \(\frac{1}{729}\)
NG
8 tháng 10 2020
\(\text{1, }\frac{27^4.9^3}{81^2}=\frac{\left(3^3\right)^4.\left(3^2\right)^3}{\left(3^4\right)^2}=\frac{3^{12}.3^6}{3^8}=3^{10}\)
\(\text{2, }\left(\frac{1}{5}\right)^{2002}.\left(-5\right)^{2000}=\frac{1}{5^{2002}}.5^{2000}=\frac{5^{2000}}{5^{2002}}=\frac{1}{5^2}=\frac{1}{5^2}\)
\(\text{3, }\frac{4^{11}.4^5}{2^{31}}=\frac{2^{22}.2^{10}}{2^{31}}=\frac{2^{32}}{2^{31}}=2\)
\(\text{4, }3^2.\frac{1}{243}.81^2.\frac{1}{3^2}=\frac{3^2.81^2}{3^5.3^2}=\frac{3^2.3^8}{3^7}=\frac{3^{10}}{3^7}=3^3=27\)
\(\text{5, }4^2.\frac{25^2}{2^3.5^2}+32.125=\frac{2^4.5^4}{2^3.5^2}+2^5.5^3=2.5^2+2^5.5^2=5^2.\left(2+2^5.5\right)=25.\left(2+32.5\right)=25.162=4050\)
5 tháng 10 2016
\(A=\frac{81^4.3^{10}.27^5:3^{12}}{3^{18}:9^3.243^2}=\frac{3^{16}.3^{10}.3^{15}:3^{12}}{3^{18}:3^6.3^{10}}=\frac{3^{29}}{3^{22}}=3^7\)
\(B=\frac{2.55^2-9.55^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{2.55^2-9.55^2.55^{19}}{25^{10}}:\frac{5\left(21.7^{14}-19.7^{14}\right)}{7.7^{15}+3.7^{15}}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}:\frac{5.7^{14}.2}{7^{15}.10}=\frac{55^2\left(55^{19}.9-2\right)}{25^{10}}.\frac{7^{15}.10}{5.7^{14}.2}\)Chịu ==
S
11 tháng 7 2019
1. \(\frac{x^7}{81}=27\Leftrightarrow x^7=2187\)
\(\Leftrightarrow x^7=3^7\Leftrightarrow x=3\)
2. \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy,...
3.\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow x^8\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^8=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)
4. \(\left(3x-1\right)^3=\frac{-8}{27}\Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{9}\)
TQ
3
24 tháng 9 2020
\(9^7+81^4-27^5\)
\(=\left(3^2\right)^7+\left(3^4\right)^4-\left(3^3\right)^5\)
\(=3^{14}+3^{16}-3^{15}\)
\(=3^{14}.\left(1+3^2-3\right)\)
\(=3^{14}.7⋮7\)
=> đpcm
\(25^{25}+5^{49}-125^{16}\)
\(=\left(5^2\right)^{25}+5^{49}-\left(5^3\right)^{16}\)
\(=5^{50}+5^{49}-5^{48}\)
\(=5^{48}.\left(5^2+5-1\right)\)
\(=5^{48}.29⋮29\)
=> đpcm
B
24 tháng 9 2020
Bài làm :
\(\text{1) }9^7+81^4-27^5\)
\(=\left(3^2\right)^7+\left(3^4\right)^4-\left(3^3\right)^5\)
\(=3^{14}+3^{16}-3^{15}\)
\(=3^{14}\left(1+3^2-3\right)\)
\(=3^{14}.7⋮7\)
=> Điều phải chứng minh
\(\text{2)}25^{25}+5^{49}-125^{16}\)
\(=\left(5^2\right)^{25}+5^{49}-\left(5^3\right)^{16}\)
\(=5^{50}+5^{49}-5^{48}\)
\(=5^{48}\left(5^2+5-1\right)\)
\(=5^{48}.29⋮29\)
=> Điều phải chứng minh
VN
3
16 tháng 8 2020
a) \(9^7+81^4-27^5=\left(3^2\right)^7+\left(3^4\right)^4-\left(3^3\right)^5\)
\(=3^{14}+3^{16}-3^{15}\)
\(=3^{14}\left(1+3^2-3\right)=3^{14}\cdot7⋮7\left(đpcm\right)\)
b) \(25^{25}+5^{49}-125^{16}=\left(5^2\right)^{25}+5^{49}-\left(5^3\right)^{16}\)
\(=5^{50}+5^{49}-5^{48}=5^{48}\left(5^2+5-1\right)\)
\(=5^{48}\cdot29⋮29\left(đpcm\right)\)
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