a: \(\Leftrightarrow2x-3=x\)

=>x=3

b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+\dfrac{5}{4}\cdot2^x=\dfrac{7}{32}\)

=>2^x=1/8

=>x=-3

c: =>2x+7=-4

=>2x=-11

=>x=-11/2

d: =>(4x-3)^2*(4x-4)(4x-2)=0

hay \(x\in\left\{\dfrac{3}{4};1;\dfrac{1}{2}\right\}\)

\(\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{27}\right)\)

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

\(\Rightarrow n=3\)

\(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\)

\(\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)

\(\Rightarrow n=4\)

a, \(\left(\dfrac{1}{3}\right)^n=\dfrac{1}{27}\Rightarrow\left(\dfrac{1}{3}\right)^n=\left(\dfrac{1}{3}\right)^3\)

Vì \(\dfrac{1}{3}\ne-1,\dfrac{1}{3}\ne0;\dfrac{1}{3}\ne1\) nên \(n=3\)

Vậy........

b, \(\left(\dfrac{3}{5}\right)^n=\dfrac{81}{625}\Rightarrow\left(\dfrac{3}{5}\right)^n=\left(\dfrac{3}{5}\right)^4\)

Vì \(\dfrac{3}{5}\ne-1,\dfrac{3}{5}\ne0;\dfrac{3}{5}\ne1\) nên \(n=4\)

Vậy..........

Chúc bạn học tốt!!!

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

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

=>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)\(\dfrac{625}{5^n}=5\)\(\Rightarrow\)\(\dfrac{5^4}{5^n}=5\)\(\Rightarrow\)\(5^n=5^5:5^4=5^1\Rightarrow n=1\)

b)\(\dfrac{\left(-2\right)^n}{16}=-32\Rightarrow\)\(\left(-2\right)^n=-32.16=\left(-1\right).2^5.2^4=\left(-1\right).2^9=\left(-2\right)^9\)

.............................\(\Rightarrow n=9\)

a. \(\dfrac{625}{5^n}=5\Rightarrow5^n=\dfrac{625}{5}=125\Rightarrow5^n=5^3\Rightarrow n=3\left(TMĐK\right)\)

Vậy n=3

b. \(\dfrac{\left(-2\right)^n}{16}=-32\Rightarrow\left(-2\right)^n=-32\cdot16=-512\Rightarrow\left(-2\right)^n=\left(-2\right)^9\Rightarrow n=9\left(TMĐK\right)\)

Vậy n=9

a

\(^{81^2=\left(3^4\right)^2}\)=\(3^8\)

..........(cần cù thì bù siêng năng chỉ có làm thì mới có ăn, mấy cái loại ko làm mà đòi có ăn thì ăn ĐẦU B**I)

NHÁ ĂN CỨT

\(\dfrac{625}{5^n}\)=5

=>\(\dfrac{5^4}{5^n}\) =5

=>\(5^4\) :\(5^n\) = 5

=>\(5^{4-n}\) =\(5^1\)

=>4\(-\)n=1

=>n=4-1

=>n=3

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