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a) \(5^n.25=125^2\)
\(\Rightarrow5^n.5^2=\left(5^3\right)^2\)
\(\Rightarrow5^n.5^2=5^6\)
\(\Rightarrow5^n=5^6:5^2\)
\(\Rightarrow5^n=5^4\)
\(\Rightarrow n=4\)
Vậy \(n=4.\)
b) \(3^n.9^2=27^3\)
\(\Rightarrow3^n.\left(3^2\right)^2=\left(3^3\right)^3\)
\(\Rightarrow3^n.3^4=3^9\)
\(\Rightarrow3^n=3^9:3^4\)
\(\Rightarrow3^n=3^5\)
\(\Rightarrow n=5\)
Vậy \(n=5.\)
c) \(2^4.4^n=8^6\)
\(\Rightarrow\left(2^2\right)^2.4^n=2^{18}\)
\(\Rightarrow4^2.4^n=\left(2^2\right)^9\)
\(\Rightarrow4^2.4^n=4^9\)
\(\Rightarrow4^n=4^9:4^2\)
\(\Rightarrow4^n=4^7\)
\(\Rightarrow n=7\)
Vậy \(n=7.\)
Chúc bạn học tốt!
a)
\(3:\left(\dfrac{9}{4}\right)=\dfrac{3}{4}:\left(6.x\right)\\ \Rightarrow3.6.x=\dfrac{3}{4}.\dfrac{9}{4}\\ x=\dfrac{3}{4}.\dfrac{9}{4}.\dfrac{1}{3}.\dfrac{1}{6}\\ x=\dfrac{3}{4.4.2}\\ x=\dfrac{3}{32}\)
b)
\(4,5:0,3=\left(5.0,09\right):\left(0,01.x\right)\\ 0,01.x.4,5=5.0,09.0,3\\ x=5.\dfrac{9}{100}.\dfrac{3}{10}.100.\dfrac{10}{45}\\ x=3\)
d)
\(\left(\dfrac{1}{9}.x\right)=\dfrac{7}{4}:\dfrac{2}{25}\\ \left(\dfrac{1}{9}.x\right)=\dfrac{7}{4}.\dfrac{25}{2}\\ x:\dfrac{7}{4}=\dfrac{25}{2}:\dfrac{1}{9}\\ x=\dfrac{25}{2}.9.\dfrac{7}{4}\\ x=\dfrac{1575}{8}\\ x=196\dfrac{7}{8}\)
e)
\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\\ -x.x=-2.\dfrac{8}{25}\\ -x^2=-\dfrac{16}{25}=-\dfrac{4^2}{5^2}\\ -x^2=-\left(\dfrac{4}{5}\right)^2\\ \Rightarrow x=\dfrac{4}{5}\)
Chúc bạn học tốt 
a: \(\dfrac{x-6}{7}+\dfrac{x-7}{8}+\dfrac{x-8}{9}=\dfrac{x-9}{10}+\dfrac{x-10}{11}+\dfrac{x-11}{12}\)
\(\Leftrightarrow\left(\dfrac{x-6}{7}+1\right)+\left(\dfrac{x-7}{8}+1\right)+\left(\dfrac{x-8}{9}+1\right)=\left(\dfrac{x-9}{10}+1\right)+\left(\dfrac{x-10}{11}+1\right)+\left(\dfrac{x-11}{12}+1\right)\)
=>x+1=0
hay x=-1
c: |x-2|=13
=>x-2=13 hoặc x-2=-13
=>x=15 hoặc x=-11
d: \(\Leftrightarrow3\left|x-2\right|+4\left|x-2\right|=2-\dfrac{1}{3}=\dfrac{5}{3}\)
=>7|x-2|=5/3
=>|x-2|=5/21
=>x-2=5/21 hoặc x-2=-5/21
=>x=47/21 hoặc x=37/21
a)\(\left(5x+1\right)^2=\frac{36}{49}\\ \left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\\ \Rightarrow\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=\frac{-6}{7}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{-1}{35}\\x=\frac{-13}{35}\end{matrix}\right.\)
vậy...
2.
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
⇒ \(5x+1=\pm\frac{6}{7}\)
⇒ \(\left[{}\begin{matrix}5x+1=\frac{6}{7}\\5x+1=-\frac{6}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x=\frac{6}{7}-1=-\frac{1}{7}\\5x=\left(-\frac{6}{7}\right)-1=-\frac{13}{7}\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-\frac{1}{7}\right):5\\x=\left(-\frac{13}{7}\right):5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{35}\\x=-\frac{13}{35}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{35};-\frac{13}{35}\right\}.\)
Chúc bạn học tốt!
1. So sánh
a) \(25^{50}\) và \(2^{300}\)
\(25^{50}=25^{1.50}=\left(25^1\right)^{50}=25^{50}\)
\(2^{300}=2^{6.50}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25< 64\) nên \(25^{50}< 64^{50}\)
Vậy \(25^{50}< 2^{300}\)
b) \(625^{15}\) và \(12^{45}\)
\(625^{15}=625^{1.15}=\left(625^1\right)^{15}=625^{15}\)
\(12^{45}=12^{3.15}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625< 1728\) nên \(625^{15}< 1728^{15}\)
Vậy \(625^{15}< 12^{45}\)
1.So sánh
a)\(25^{50}\) và \(2^{300}\)
Ta có : \(2^{300}=\left(2^6\right)^{50}=64^{50}\)
Vì \(25^{50}< 64^{50}\) nên \(25^{50}< 2^{300}\)
b)\(625^{15}\) và \(12^{45}\)
Ta có : \(12^{45}=\left(12^3\right)^{15}=1728^{15}\)
Vì \(625^{15}< 1728^{15}\) nên \(625^{15}< 12^{45}\)
B1. phân a tui ko bt nha :>
\(B=\frac{2^{13}\cdot9^4}{6^6\cdot8^3}\)
\(=\frac{2^{13}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}\)
\(=\frac{2^{13}\cdot3^8}{2^6\cdot3^6\cdot2^9}\)
\(=\frac{2^{13}\cdot3^8}{2^{15}\cdot3^6}\)
\(=\frac{1\cdot3^2}{2^2\cdot1}\)
\(=\frac{1\cdot9}{4\cdot1}\)
\(=\frac{9}{4}\)