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\(25^6\cdot8^4\)
\(=\left(5^2\right)^6\cdot\left(2^3\right)^4\)
\(=5^{2\cdot6}\cdot2^{3\cdot4}\)
\(=5^{12}\cdot2^{12}\)
\(=\left(5\cdot2\right)^{12}\)
\(=10^{12}\)
\(25^6.8^4\)
\(=\left(5^2\right)^6.\left(2^3\right)^4\)
\(=5^{2.6}.2^{3.4}\)
\(=5^{12}.2^{12}\)
\(=\left(5.2\right)^{12}\)
\(=10^{12}\)
Ta có: +) \({({2^2})^3} = {2^2}{.2^2}{.2^2} = {2^{2 + 2 + 2}} = {2^6}\)
+) \({\left[ {{{( - 3)}^2}} \right]^2} = {( - 3)^2}.{( - 3)^2} = {( - 3)^{2 + 2}} = {( - 3)^4}\)
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}`
a) 2.2.2.2 = \({2^4}\). Cơ số 2, số mũ 4
b) 5.5.5 = \({5^3}\). Cơ số 5, số mũ 3
\(\left(0,0625\right)^2=\left(0,5^4\right)^2\)
\(=0,5^{4\cdot2}\)
\(=0,5^8\)
\(\left(0,0625\right)^2=\left[\left(0,5\right)^4\right]^2=\left(0,5\right)^8\)
\(\left(\dfrac{1}{27}\right)^5\) = \(\left(\dfrac{1}{3^3}\right)^5\) = \(\left(\dfrac{1}{3}\right)^{15}\)
\(\left(\dfrac{1}{27}\right)^5=\left[\left(\dfrac{1}{3}\right)^3\right]^5=\dfrac{1}{3}^{3.5}=\dfrac{1}{3}^{15}\)
\(25^6.8^4\)
\(=\left(5^2\right)^6.\left(2^3\right)^4\)
\(=5^{12}.2^{12}\)
\(=\left(5.2\right)^{12}=10^{12}\)
= 10\(^{12}\)