=9.9..1/81.27

=81.1/81.27

=1.27

=27

9.3^2.1/81.27

=9.9.1/81.27

=81.1/81.27

=1.27

=27

Đề có phải là thế này ko

9.3^9.\(\frac{1}{81}\)

a,9<3^x<3⁹.3^-2

\(\Leftrightarrow3^2< 3< 3^7\)

\(\Rightarrow S=\left\{3;4;5;6\right\}\)

a) \(2.4.16.32.2^4=2.2^2.2^4.2^5.2^4=2^{16}\)

b) \(\left(4.2^5\right):\left(2^3.\frac{1}{16}\right)=\left(2^2.2^5\right):\left(2^3.\left(\frac{1}{2}\right)^4\right)=2^7:\frac{1}{2}=2^8\)

c) \(9.3^3.\frac{1}{81}.27=3^2.3^3.\left(\frac{1}{3}\right)^4.3^3=3^4\)

d)\(2^2.4.\frac{32}{2^2}.2^5=2^2.2^2.2^3.2^5=2^{12}\)

a/\(2.4.16.32.2^4=2.2^2.2^4.2^5.2^4=2^{16}\)

b/\(9.3^3.\dfrac{1}{81}=3^2.3^3.3^{-4}=3\)

a) \(2\cdot4\cdot16\cdot32\cdot2^4\)

\(=2^1\cdot2^2\cdot2^4\cdot2^5\cdot2^4\)

\(=2^{16}\)

b) \(9\cdot3^3\cdot\dfrac{1}{81}\cdot27\)

\(=3^2\cdot3^3\cdot3^{-4}\cdot3^3\)

\(=3^4\)

a) \(9\cdot3^2\cdot\dfrac{1}{81}=\dfrac{9\cdot9}{81}=\dfrac{81}{81}=1\)

b) \(2\dfrac{1}{2}+\dfrac{4}{7}:\left(-\dfrac{8}{9}\right)\)\(=\dfrac{5}{2}+\dfrac{4}{7}\cdot\left(-\dfrac{9}{8}\right)\)

\(=\dfrac{5}{2}-\dfrac{9}{14}\)\(=\dfrac{35}{14}-\dfrac{9}{14}=\dfrac{26}{14}=\dfrac{13}{7}\)

c) \(3,75\cdot7,2+2,8\cdot3,75=3,75\left(7,2+2,8\right)=3,75\cdot10=37,5\)

\(3^x+9\cdot3^{-x}=10\)

=>\(3^x+\dfrac{9}{3^x}=10\)

=>\(\dfrac{\left(3^x\right)^2+9}{3^x}=10\)

=>\(\left(3^x\right)^2+9=10\cdot3^x\)

=>\(\left(3^x\right)^2-10\cdot3^x+9=0\)

=>\(\left(3^x-1\right)\left(3^x-9\right)=0\)

=>\(\left[{}\begin{matrix}3^x-1=0\\3^x-9=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(\frac{2^2\cdot4\cdot32}{\left(-2\right)^2\cdot2^5}=\frac{2^2\cdot4\cdot2^5}{4\cdot2^5}=2^2=4\)

\(9\cdot3^2\cdot\frac{1}{81}\cdot27=9\cdot9\cdot\frac{1}{81}\cdot27=81\cdot\frac{1}{81}\cdot27=27\)

đổi thành thùa số nguyên tố là tính được nha bn

giúp mk vs

A)4x2^5:(2^3.1/16)=2^2.2^5:(8.1/16)=2^7:1/2=2^7.2=2^8

B)3^2.2^5.(2/3)^2=3^2.2^5.4/9=2^5.9.4/9=2^5.4=2^5.2^2=2^7

C) 9.3^3.1/81.3^2=3^3.9.9.1/81=3^3.81.1/81=3^3