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\(\frac{3^6.45^5-15^{13}.5^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3^2.5\right)^5-\left(3.5\right)^{13}.5^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3^2.5\right)^6}=\frac{3^{16}.5^5-3^{13}.5^4}{3^{12}.5^6+3^{12}.5^6}\)
\(\frac{3^{13}.5^4\left(3^3.5-1\right)}{2.3^{12}.5^6}=\frac{3.44}{2.5^2}=\frac{3.22}{5^2}=\frac{66}{25}\)
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a) \(\frac{7^3.5^8}{49.25^4}=\frac{7^3.5^8}{7^2.5^8}=7\)
b) \(\frac{3^9.25.5^3}{15.625.3^8}=\frac{3^9.5^2.5^3}{3.5.5^4.3^8}=\frac{3^9.5^5}{3^9.5^5}=1\)
c) \(\frac{2^{50}.3^{61}+2^{90}.3^{16}}{2^{51}.3^{61}+2^{91}.3^{16}}=\frac{2^{50}.3^{16}\left(3^{45}+2^{40}\right)}{2^{51}.3^{16}\left(3^{45}+2^{40}\right)}=\frac{1}{2}\)
d) \(\left(\frac{2}{5}-\frac{1}{2}\right)^2+\left(\frac{1}{2}+\frac{3}{5}\right)^2\)
\(=\left(\frac{-1}{10}\right)^2+\left(\frac{11}{10}\right)^2\)
\(=\frac{1}{100}+\frac{121}{100}=\frac{122}{100}=\frac{61}{50}\)
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\(\frac{3^6.45^4-15^{13}.9^{-9}}{27^4.25^3+45^6}=\frac{3^6.\left(3.3.5\right)^4-\left(3.5\right)^{13}-\left(3^2\right)^{-9}}{\left(3^3\right)^4.\left(5^2\right)^3+\left(3.15\right)^6}=\frac{3^6.3^4.3^4.5^4-3^{13}.5^{13}-3^{-18}}{3^{12}.5^6+3^6.15^6}=\frac{3^{14}.5^4-3^{13}.5^{13}-3^{-18}}{3^{12}.5^6+3^6.\left(3.5\right)^6}=\frac{ }{ }\)
Mình nghĩ hết được rồi
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a. \(\Rightarrow5^{-1}.5^{2n}=5^3\)
\(\Rightarrow5^{2n-1}=5^3\)
=> 2n-1=3
=> 2n=4
=> n=2
b. \(\Rightarrow3^{n-1}+6.3^{n-1}=7.3^6\)
\(\Rightarrow\left(1+6\right).3^{n-1}=7.3^6\)
\(\Rightarrow7.3^{n-1}=7.3^6\)
=> n-1=6
=> n=7
c. \(\Rightarrow3^4<3^{-2}.3^{3n}<3^{10}\)
\(\Rightarrow3^4<3^{3n-2}<3^{10}\)
\(\Rightarrow3n-2\in\left\{5;6;7;8;9\right\}\)
\(\Rightarrow3n\in\left\{7;8;9;10;11\right\}\)
\(\text{Mà n là số nguyên}\Rightarrow n=3\).
d. \(\Rightarrow5^2<5^{n-1}<5^4\)
\(\Rightarrow n-1=3\)
\(\Rightarrow n=4\).
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\(\dfrac{45^{10}\cdot5^{20}}{75^{15}}=\dfrac{\left(3^2\cdot5\right)^{10}\cdot5^{20}}{\left(3\cdot5^2\right)^{15}}=\dfrac{3^{20}\cdot5^{10}\cdot5^{20}}{3^{15}\cdot5^{30}}=3^5=243\\ \dfrac{6^6+6^3+3^3+3^6}{-73}=\dfrac{46656+216+27+729}{-73}=-\dfrac{47628}{73}\\ \dfrac{27^7+3^{15}}{9^9-27}=\dfrac{\left(3^3\right)^7+3^{15}}{\left(3^2\right)^9-3^3}=\dfrac{3^{21}+3^{15}}{3^{18}-3^3}=\dfrac{3^{15}\left(3^6+1\right)}{3^3\left(3^{15}-1\right)}=\dfrac{3^5\cdot730}{3^{15}-1}\\ \dfrac{8^{20}+4^{20}}{4^{25}+64^5}=\dfrac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\dfrac{2^{60}+2^{40}}{2^{50}+2^{30}}=\dfrac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
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