a: =>9^n=9

=>n=1

b: =>5^n=5

=>n=1

c: \(\Leftrightarrow\left(-27\right)^n=-243\)

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

=>3n=5

=>n=5/3

d: =>2^n*9/2=9*2^5

=>2^n=9*2^5:9/2=2^5*2=2^6

=>n=6

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)x ∈ ∅

b) x=3

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

a) \(3^3\)

b)\(2^8\)

c) \(2^7\)

d) \(3^1\)

a) 9.3³.\(\dfrac{1}{81}\) .3² = 3². 3³.\(\dfrac{1}{3^4}\) . 3² = 3³

b) 4. 2⁵: \(\) (2³.\(\dfrac{1}{16}\))= 2². 2⁵: 2³. \(\dfrac{1}{2^4}\) = 2⁷: \(\dfrac{1}{2}\) = 2⁷. 2= 2⁸

c) 3². 2⁵. \(\left(\dfrac{2}{3}\right)^2\) = 3². 2⁵. \(\dfrac{2^2}{3^2}\) = 2⁵. 2² = 2⁷

d) \(\left(\dfrac{1}{3}\right)^2\) .\(\dfrac{1}{3}\) . 9² = \(\dfrac{1}{9}.\dfrac{1}{3}\). 9² = \(\dfrac{9}{3}\) = 3¹

a không bít !

b chịu

c quên rồi

d . . .

a) 27^n : 3^n = 9

     (27 : 3)^n  = 9

         9^n        = 9

=> n = 1

b) 25/5^n = 5

       5^n     = 25 : 5

       5^n     = 5

=> n = 1

c) 81/(-3)^n = -243

         (-3)^n   = -243 : 81

         (-3)^n   = -3

=> n = 1

d) 1/2 . 2^n + 4 . 2^n = 9 . 2^5

       2^n . (1/2 + 4)     = 9 . 32

       2^n . 9/2              = 288

       2^n                       = 288 : 9/2

       2^n                       = 64

       2^n                       = 2^6

=> n = 6

thank you for doing your homework well !

\(a)27^n\div3^n=9\)

\(\Rightarrow\left(27\div3\right)^n=9\)

\(\Rightarrow9^n=9\Rightarrow n=1\)

\(b)\frac{25}{5^n}=5\Rightarrow25=5^{n+1}\Rightarrow5^2=5^{n+1}\Rightarrow2=n+1\Rightarrow n=1\)

\(c)\frac{81}{\left(-3\right)^n}=-243\Rightarrow\frac{81}{\left(-3\right)^n}=\left(-3\right)^5\)

\(\Rightarrow81=\left(-3\right)^{n+5}\Rightarrow\left(-3\right)^4=\left(-3\right)^{n+5}\)

\(\Rightarrow4=n+5\Rightarrow n=-1\)

\(d)\frac{1}{2}.2^n+4.2^n=9.2^5\Rightarrow2^n\left(\frac{1}{2}+4\right)=9.2^5\)

\(\Rightarrow2^n.\frac{9}{2}=9.2^5\Rightarrow2^n=9.2^5.\frac{2}{9}\Rightarrow2^n=\frac{9.2^5.2}{9}=2^6\)

\(\Rightarrow n=6\)

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

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

   \(2^n.\frac{9}{2}=3^2.5^n\)

Đây là tính hợp lí ... mà câu a là 27,5 chứ không phải 2,75...

\(A=\dfrac{7,5-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{27,5-2,2+\dfrac{11}{7}+\dfrac{11}{3}}=\dfrac{\dfrac{15}{2}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{55}{2}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{3}}\\ =\dfrac{3\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{5}{2}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)

b: \(=26:\left[\dfrac{3:0.1}{2.5\cdot2}+\dfrac{0.25\cdot4}{2}\right]+\dfrac{2}{3}\cdot\dfrac{21}{4}\)

\(=26:\left[\dfrac{30}{5}+1\right]+\dfrac{42}{12}\)

\(=\dfrac{26}{7}+\dfrac{42}{12}=\dfrac{101}{14}\)

c: \(=\left[\dfrac{4-3}{386}\cdot\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\dfrac{25}{4002}\cdot\dfrac{2001}{25}+\dfrac{9}{2}\right]\)

\(=\dfrac{\left(\dfrac{1}{34}+\dfrac{33}{34}\right)}{\dfrac{1}{2}+\dfrac{9}{2}}=1:5=\dfrac{1}{5}\)

i love you

a) \(\dfrac{790^4}{79^4}\) = \(\dfrac{79^4.10^4}{79^4}\) = 10⁴ = 10000

b) \(\dfrac{3^2}{0,375^2}\) = \(\dfrac{0,375^2.8^2}{0,375^2}\) = 8² = 64

c) \(3^2.\dfrac{1}{243}.81^2.\dfrac{1}{3^3}\) = 3² . 3^-5 . 3⁸ . 3^-3 = 3² = 9

d) \(\left(4.2^5\right):\left(2^3.\dfrac{1}{16}\right)\) = 2⁷ : (2³ . 2^-4) = 2⁷ : 2^-1 = 2⁷ : 1/2 = 2⁸