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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(0,09\right)^3=\left(\dfrac{9}{100}\right)^3=\left[\left(\dfrac{3}{10}\right)^2\right]^3=\left(\dfrac{3}{10}\right)^6\\ \left(\dfrac{3}{10}\right)^8=\left(\dfrac{3}{10}\right)^8\\ \left(0,027\right)^2=\left(\dfrac{27}{1000}\right)^2=\left[\left(\dfrac{3}{10}\right)^3\right]^2=\left(\dfrac{3}{10}\right)^6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2^{225}=2^{3.75}=\left(2^3\right)^{75}=8^{75}\)
\(3^{150}=3^{2.75}=\left(3^2\right)^{75}=9^{75}\)
Vì \(8^{75}\)\(< \)\(9^{75}\) nên \(2^{225}< 3^{150}\)
b) \(2^{12}=16^3\)
\(4^{18}=16^9\)
K mk nha mk nhanh nhất 100% đấy
a)Ta có: 2225 = (23)75 và 3150 = (32)75
Do 23 < 32 => (23)75 < (33)75 => 2225 < 3150
b) 212 = (24)3 = 163
418 = (42)9 = 169
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: +) \({({2^2})^3} = {2^2}{.2^2}{.2^2} = {2^{2 + 2 + 2}} = {2^6}\)
+) \({\left[ {{{( - 3)}^2}} \right]^2} = {( - 3)^2}.{( - 3)^2} = {( - 3)^{2 + 2}} = {( - 3)^4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\begin{array}{l}{\left( {\frac{1}{4}} \right)^8} = {[{\left( {\frac{1}{2}} \right)^2}]^8} = {(\frac{1}{2})^{2.8}} = {(\frac{1}{2})^{16}};\\{\left( {\frac{1}{8}} \right)^3} = {[{(\frac{1}{2})^3}]^3} = {(\frac{1}{2})^{3.3}} = {(\frac{1}{2})^9}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4^{18}=\left(2^2\right)^{18}=2^{36}=\left(2^3\right)^{12}=8^{12}\)
\(2^{12}=2^{3.4}=\left(2^3\right)^4=8^4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 0,097
0,0273=(0,093)3=0,099
=> 0,097 và 0,099 cùng cơ số là 0,09
a) 8 = 23
425 = 25.35.75
16 = 24
b) (0,09)3 = (3/10)6
(3/10)8 = (3/10)8
0,027 = (3/10)3
`@` `\text {Ans}`
`\downarrow`
`a)`
`8 = 2^3`
`32^5` chứ ạ?
`32^5 = (2^5)^5 = 2^10`
`16 = 2^4`
`b)`
`(0,09)^3 = (0,3^2)^3 = 0,3^6` hay `(3/10)^6`
`(3/10)^8 = (3/10)^8`
`(0,027) = (0,3)^3` hay `(3/10)^3`
`@` `\text {Kaizuu lv uuu}`