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\(8;a,3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)
\(=\frac{3^2.\left(3^4\right)^2}{243.3^3}\)
\(=\frac{3^2.3^8}{3^5.3^3}\)
\(=\frac{3^{10}}{3^8}=3^2=9\)
\(b,\frac{4.2^5}{2^3.\frac{1}{16}}\)
\(=\frac{2^2.2^5}{2^3.\frac{1}{2^4}}\)
\(=\frac{2^7}{\frac{1}{2}}=2^7.2=2^8\)
a, \(3^2.\frac{1}{243}.81^2.\frac{1}{3}^3\)
\(=3^2.\frac{1}{243}.\left(3^4\right)^2.\frac{1}{27}\)
\(=3^2.\frac{1}{243}.3^8.\frac{1}{27}\)
\(=\frac{3^2.3^8}{243.27}\)
\(=\frac{3^2.3^8}{3^5.3^3}\)
\(=\frac{3^{10}}{3^8}=3^2=9\)
b, \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)\)
\(=\left(2^2.2^5\right):\left(8.\frac{1}{16}\right)\)
\(=2^7:\frac{1}{2}\)
\(=2^8\)
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\(1-\frac{2}{3}-\frac{2}{9}-\frac{2}{27}-\frac{2}{81}-\frac{2}{243}\)
\(=\frac{243}{243}-\frac{162}{243}-\frac{54}{243}-\frac{6}{243}-\frac{2}{243}=\frac{1}{243}\)
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3^ x -1 = 1/243
3^x =1/243 +1
3^x = 244 / 243
Ta thấy đây ko phải lũy thừa của 3 => Ko có x thỏa mãn
81^-2x . 27^x =9^5
81^-2 . 81^x . 27^x =9^5
1/9^4 . (81.27)^x =9 ^5
3^6x = 9^5 : 1/9^4
3^6x = 9^9
3^6x = 3^18
=> 6x =18
x=3
2^x +2^x +3 =144
2.(2^x) =141
2^x+1 = 141
Ta thấy 141 ko phải lũy thừa của 2 => ko có x thỏa mãn
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\(\left(-3\right).\frac{1}{243}.81^2.3^{\left(-3\right)}\)
\(=\left(-3\right).\frac{1}{3^3}.\frac{1}{3^5}.3^8\)
\(=\frac{-1}{3^2}.3^3\)
\(=-3\)
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1/81 = (1/9)^2
243 = 3^5
8. 3^3 = 2^3 . 3^3 = 6^3
81.2^8 = 9^2 . (2^4)^2 = 9^2 . 16^2 = ( 9. 16)^2 = 144^2
`@` `\text {Ans}`
`\downarrow`
\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^3}\)
`=`\(\dfrac{3^2}{243}\cdot\dfrac{81^2}{3^3}\)
`=`\(\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^3}=\dfrac{1}{3^3}\cdot3^5=\dfrac{3^5}{3^3}=3^2=9\)
\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^2}\)
\(=3^2\cdot\dfrac{1}{3^5}\cdot\left(3^4\right)^2\cdot\dfrac{1}{3^2}\)
\(=3^2\cdot\dfrac{1}{3^5}\cdot3^8\cdot\dfrac{1}{3^2}\)
\(=\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^2}\)
\(=\dfrac{3^2}{3^5}\cdot3^6\)
\(=\dfrac{3^2\cdot3^6}{3^5}\)
\(=3^2\cdot3\)
\(=3^3\)
\(=27\)