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 \(x-1=4\Rightarrow x=5\)

a)\(\left(\dfrac{1}{2}\right)^n=\dfrac{1}{32}\)

=>\(\left(\dfrac{1}{2}\right)^n=\left(\dfrac{1}{2}\right)^5\)

=>n=5

b)\(\left(\dfrac{343}{125}\right)=\left(\dfrac{7}{5}\right)^n\)

=>\(\left(\dfrac{7}{5}\right)^3=\left(\dfrac{7}{5}\right)^n\)

=>n=3

c)\(\dfrac{16}{2^n}=2\)

=>2ⁿ=\(\dfrac{16}{2}\)

=>2ⁿ=8

=>2ⁿ=2³

=>n=3

d)\(\dfrac{\left(-3\right)^n}{81}=-27\)

=>(-3)ⁿ=-27.81

=>(-3)ⁿ=-2187

=>(-3)ⁿ=(-3)⁷

=>n=7

e)8ⁿ:2ⁿ=4

=>(2³)ⁿ:2ⁿ=4

=>2³ⁿ:2ⁿ=4

=>2^3n-n=4

=>2²ⁿ=4

=>2²ⁿ=2²

=>2n=2

=>n=1

f)3².3ⁿ=3⁵

=>3ⁿ=3⁵:3²

=>3ⁿ=3^5-2

=>3ⁿ=3³

=>n=3

g) (2²:4).2ⁿ=4

=>1.2ⁿ=2²

=>n=2

h)3^-2.3⁴.3ⁿ=3⁷

=>\(\left(\dfrac{1}{3}\right)^2\).3⁴.3ⁿ=3⁷

=>3².3ⁿ=3⁷

=>3²⁺ⁿ=3⁷

=>2+n=7

=>n=5

a)x ∈ ∅

b) x=3

\(a,\Rightarrow2^3< 2^x\le2^4\Rightarrow x=4\\ b,\Rightarrow3^3< 3^{12}:3^x< 3^5\\ \Rightarrow3^3< 3^{12-x}< 3^5\\ \Rightarrow12-x=4\Rightarrow x=8\)

1. Tìm n, biết:

a) \(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)

\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)

(-2)^{n + 2} = (-2)⁵

n + 2 = 5

n = 5 - 2

n = 3.

b) \(\dfrac{8}{2^n}=2\)

\(\Rightarrow\dfrac{2^3}{2^n}=2\)

\(\Rightarrow\) 2ⁿ . 2 = 2³

n + 1 = 3

n = 3 - 1

n = 2.

c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)

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

2n - 1 = 3

2n = 3 + 1

2n = 4

n = 4 : 2

n = 2.

2. Tính:

a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)

\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)

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

\(=\left(\dfrac{1}{2}\right)^7\)

\(=\dfrac{1}{128}\)

b) 27³ : 9³

= (3³)³ : (3²)³

= 3⁹ : 3⁶

= 3³

= 27

c) 125² : 25³

= (5³)² : (5²)³

= 5⁶ : 5⁶

= 1

d) \(\dfrac{27^2.8^5}{6^6.32^3}\)

\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)

\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)

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

\(=\dfrac{1}{64}.\)

B2 :

b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)

c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1

a, \(\dfrac{16}{2^n}=2\)

\(2^n=\dfrac{16}{2}\)

\(2^n=8\)

\(2^n=2^3\)

=> n = 3

b, \(\dfrac{\left(-3\right)^n}{81}=-27\)

\(\left(-3\right)^n=-27\cdot81\)

\(\left(-3\right)^n=\left(-3\right)^3\cdot3^4\)

\(\left(-3\right)^n=\left(-3\right)^7\)

=> n = 7

c, \(8^n:2^n=4\)

\(2^{3n}:2^n=2^2\)

\(2^{2n}=2^2\)

=> 2n = 2

n = 2:2

n = 1

a )

\(\dfrac{16}{2^n}=2\) \(\Leftrightarrow16:x=2\)

\(\Rightarrow x=8\)

\(2^n=8\Rightarrow n=3\)

b )

\(\dfrac{\left(-3\right)^n}{81}=-27\) \(\Leftrightarrow x=-27.81\)

\(\Rightarrow x=-2187\)

\(\left(-3\right)^n=-2187\Rightarrow n=7\)

c )

\(8^n:2^n=4\Leftrightarrow4^n=4\)

\(\Rightarrow n=1\)

(2^2:4)*2^n=4

3^2*3^4*3^n=3^7

a)

b)

c)

a) $\frac{16}{2^n}=\mathrm{2\; =>\; \backslash (\backslash dfrac\{2^4\}\{2^n\}=2\backslash Rightarrow2^\{4-n\}=2\backslash Rightarrow4-n=1\backslash Rightarrow\; n=3\backslash )}$

b,
\(\dfrac{\left(-3\right)^n}{81}=-27\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^4}=-27\Rightarrow\left(-3\right)^{n-4}=\left(-3\right)^3\Rightarrow n-4=3\Rightarrow n=7\)

c,\(8^n:2^n=4\Rightarrow4^n=4\Rightarrow n=1\)

=> (-3)n-4 = (-3)3

=> n - 4 = 3 => n = 7

c) 8n : 2n = 4

4n = 4.

a. x=5/6

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

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

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

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

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

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

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

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

\(a. \)

\(\dfrac{-32}{\left(-2\right)^n}=4\)

\(\Rightarrow\left(-2\right)^n=-32:4=-8\)

\(\Rightarrow\left(-2\right)^n=\left(-2\right)^3\)

\(\Rightarrow n=3\)

\(b.\)

\(\dfrac{8}{2^n}=2\)

\(\Rightarrow2^n=4\)

\(\Rightarrow2^n=2^2\)

\(\Rightarrow n=2\)

\(c.\)

\(\dfrac{16}{\left(-2\right)^n}=-8\)

\(\Rightarrow\left(-2\right)^n=-2\)

\(\Rightarrow n=1\)

a,

\(\left(4x-\dfrac{1}{3}\right)^6=1\\ \Rightarrow\left[{}\begin{matrix}4x-\dfrac{1}{3}=1\\4x-\dfrac{1}{3}=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{-2}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-1}{6}\end{matrix}\right.\)

b,

\(\left(5x-\dfrac{2}{3}\right)^2=0\\ \Rightarrow5x-\dfrac{2}{3}=0\\ 5x=\dfrac{2}{3}\\ x=\dfrac{2}{15}\)

c,

\(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\\ \Rightarrow\dfrac{1}{3}x-\dfrac{1}{2}=-2\\ \dfrac{1}{3}x=\dfrac{-3}{2}\\ x=\dfrac{-9}{2}\)

d,

\(\dfrac{81}{3^n}=3\\ \Leftrightarrow3^4:3^n=3^1\\\Leftrightarrow3^{4-n}=3^1 \\ \Rightarrow n=3\)

e,

\(\dfrac{\left(-2\right)^x}{64}=-2\\ \Leftrightarrow\left(-2\right)^x:\left(-2\right)^6=\left(-2\right)^1\\ \Leftrightarrow\left(-2\right)^{x-6}=\left(-2\right)^1\\ \Rightarrow x=7\)

f,

\(\left(-20\right)^n:10^n=16\\ \left[\left(-20\right):10\right]^n=16\\ \left(-2\right)^n=\left(-2\right)^4\\ \Rightarrow n=4\)

Bài 1:

a) \(\left(4x-\dfrac{1}{3}\right)^6=1\)

\(\Rightarrow4x-\dfrac{1}{3}=1\)

\(4x=1+\dfrac{1}{3}\)

\(4x=\dfrac{4}{3}\)

\(x=\dfrac{4}{3}:4\)

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

b) \(\left(5x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow5x-\dfrac{2}{3}=0\)

\(5x=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}:5\)

\(x=\dfrac{2}{15}\)

c) \(\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)^3=-8\)

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

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

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

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

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

\(x=\dfrac{-9}{2}\)

d) \(\dfrac{81}{3^n}=3\)

\(\Rightarrow\dfrac{3^4}{3^n}=3\)

\(\Rightarrow3^n.3=3^4\)

\(3^{n+1}=3^4\)

n + 1 = 4

n = 4 - 1

n = 3

e) \(\dfrac{\left(-2\right)^x}{64}=-2\)

\(\Rightarrow\dfrac{\left(-2\right)^x}{\left(-2\right)^6}=-2\)

\(\Rightarrow\left(-2\right)^x=\left(-2\right)^6.\left(-2\right)\)

\(\left(-2\right)^x=\left(-2\right)^7\)

x = 7

f) (-20)ⁿ : 10ⁿ = 16

(-20 : 10)ⁿ = 16

(-2)ⁿ = 16

(-2)ⁿ = (-2)⁴

n = 4.

Sửa lại câu a) \(\left(4x-\dfrac{1}{3}\right)^6=1\)

\(\Rightarrow\left(4x-\dfrac{1}{3}\right)^6=\left(\pm1\right)^6\)

\(\Rightarrow\left\{{}\begin{matrix}4x-\dfrac{1}{3}=1\\4x-\dfrac{1}{3}=-1\end{matrix}\right.\left\{{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{-2}{3}\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{-1}{6}\end{matrix}\right.\)