a. 96 : 8 : 2 = 6

96 : ( 8 x 2 ) = 6

96 : 8 : 2 = 96 : ( 8 x 2 )

b. 81 : 9 : 3 = 3

81 : ( 9 x 3 ) = 3

81 : 9 : 3 = 81 : ( 9x 3)

a)27ⁿ:3ⁿ=9

   (27:3)ⁿ=9

   9ⁿ=9¹

   n=1

Vậy n=1

b)\(\left(\frac{25}{5}\right)^n=5\)

    \(5^n=5^1\)

    n=1

Vạy n=1

c)\(\left(-\frac{81}{3}\right)^n=-243\)

   \(\left(-27\right)^n=\left(-3\right)^5\)

   \(\left[\left(-3\right)^3\right]^n=\left(-3\right)^5\)

   \(\left(-3\right)^{3n}=\left(-3\right)^5\)

    \(3n=5\)

   \(n=\frac{5}{3}\)

Vậy \(n=\frac{5}{3}\)

d)\(\frac{1}{2}.2^n+4.2^n=9.5^n\)

   \(2^n.\left(\frac{1}{2}+4\right)=9.5^n\)

   \(2^n.\frac{9}{2}=3^2.5^n\)

câu 1: A={0;1;2;3}

Câu 2: A={13;14;15}

Câu 3:E={1;2;3;4}

Câu 4: 2⁵

Câu 5:5

Câu 6: 6²=36

4³=64

Vậy 4³ lớn hơn

😮😮😮😮😮Hihi

a. x=5/6

\(a,\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}=\left(\dfrac{1}{3}\right)^3\\ \Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\Rightarrow x=\dfrac{5}{6}\\ b,\Rightarrow\left(\dfrac{3}{2}\right)^{2x-1}:\left(\dfrac{3}{2}\right)^9=\left(\dfrac{3}{2}\right)^4\\ \Rightarrow2x-1-9=4\\ \Rightarrow2x=14\Rightarrow x=7\\ c,\Rightarrow2^{x-1}+2^{x+2}=9\cdot2^5\\ \Rightarrow2^{x-1}\left(1+2^3\right)=9\cdot2^5\\ \Rightarrow2^{x-1}\cdot9=9\cdot2^5\\ \Rightarrow2^{x-1}=2^5\Rightarrow x-1=5\Rightarrow x=6\\ d,\Rightarrow\left(2x+1\right)^2=12+69=81\\ \Rightarrow\left[{}\begin{matrix}2x+1=9\\2x+1=-9\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-5\end{matrix}\right.\)

\(a,\dfrac{2}{3}+\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{19}{27}-\dfrac{2}{3}\)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\)

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(\dfrac{1}{3}\right)^3\)

\(\Rightarrow x-\dfrac{1}{2}=\dfrac{1}{3}\)

\(x=\dfrac{1}{2}+\dfrac{1}{3}\)

\(x=\dfrac{1}{5}\)

Bài 1:

a, \(\left(x-2\right)^2=9\)

\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)

b, \(\left(3x-1\right)^3=-8\)

\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)

\(\Rightarrow x=-\dfrac{1}{3}\)

c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)

\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)

\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)

d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)

\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)

Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)

e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)

\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)

Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)