`@` `\text {Ans}`

`\downarrow`

\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^3}\)

`=`\(\dfrac{3^2}{243}\cdot\dfrac{81^2}{3^3}\)

`=`\(\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^3}=\dfrac{1}{3^3}\cdot3^5=\dfrac{3^5}{3^3}=3^2=9\)

\(3^2\cdot\dfrac{1}{243}\cdot81^2\cdot\dfrac{1}{3^2}\)

\(=3^2\cdot\dfrac{1}{3^5}\cdot\left(3^4\right)^2\cdot\dfrac{1}{3^2}\)

\(=3^2\cdot\dfrac{1}{3^5}\cdot3^8\cdot\dfrac{1}{3^2}\)

\(=\dfrac{3^2}{3^5}\cdot\dfrac{3^8}{3^2}\)

\(=\dfrac{3^2}{3^5}\cdot3^6\)

\(=\dfrac{3^2\cdot3^6}{3^5}\)

\(=3^2\cdot3\)

\(=3^3\)

\(=27\)

\(=\dfrac{3^2.81^2}{243.3^3}=\dfrac{3^2.3^4}{3^5.3^3}=\dfrac{3^6}{3^8}=\dfrac{1}{3^2}=\dfrac{1}{9}\)

\(=3^2.\frac{1}{3^5}.3^8.\frac{1}{3^3}\)

\(=9\)

\(8;a,3^2.\frac{1}{243}.81^2.\frac{1}{3^3}\)

\(=\frac{3^2.\left(3^4\right)^2}{243.3^3}\)

\(=\frac{3^2.3^8}{3^5.3^3}\)

\(=\frac{3^{10}}{3^8}=3^2=9\)

\(b,\frac{4.2^5}{2^3.\frac{1}{16}}\)

\(=\frac{2^2.2^5}{2^3.\frac{1}{2^4}}\)

\(=\frac{2^7}{\frac{1}{2}}=2^7.2=2^8\)

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, \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)\)

\(=\left(2^2.2^5\right):\left(8.\frac{1}{16}\right)\)

\(=2^7:\frac{1}{2}\)

\(=2^8\)

=1/243 nha bn có j k mk nka 

\(1-\frac{2}{3}-\frac{2}{9}-\frac{2}{27}-\frac{2}{81}-\frac{2}{243}\)

\(=\frac{243}{243}-\frac{162}{243}-\frac{54}{243}-\frac{6}{243}-\frac{2}{243}=\frac{1}{243}\)

Nông Hà Thanh

\(3^2.\frac{1}{243}.81^2.\frac{1}{3^2}\)

\(=\frac{3^2}{3^5}.\frac{81^2}{3^2}\)

\(=\frac{1}{3^3}.27^2\)

\(=\frac{27^2}{3^3}\)

\(=\frac{3^6}{3^3}\)

\(=3^2\)

\(=9\)

=3^2*1/3^5*3^4/2*1/3^2

=1/6

Em cảm ơn

a: \(2^{x^2-1}=256\)

=>\(2^{x^2-1}=2^8\)

=>\(x^2-1=8\)

=>\(x^2=9\)

=>\(x\in\left\{3;-3\right\}\)

b: \(3^{x^2+3x}=81\)

=>\(3^{x^2+3x}=3^4\)

=>\(x^2+3x=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=1\end{matrix}\right.\)

c: \(2^{x^2-5x}=64\)

=>\(2^{x^2-5x}=2^6\)

=>\(x^2-5x=6\)

=>\(x^2-5x-6=0\)

=>(x-6)(x+1)=0

=>\(\left[{}\begin{matrix}x-6=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^x=243\)

=>\(\left(\dfrac{1}{3}\right)^x=3^5=\left(\dfrac{1}{3}\right)^{-5}\)

=>x=-5

e: \(\left(\dfrac{1}{3}\right)^{x+5}=3^{2x+1}\)

=>\(3^{-x-5}=3^{2x+1}\)

=>-x-5=2x+1

=>-3x=6

=>x=-2