Question

Tìm n biết

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

b) \(\frac{8}{2^n}\)= 2

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

Answer 1

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

\(\frac{\left(-2\right)^5}{\left(-2\right)^n}=4\)

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

=> 5-n = 2

n = 3

b) \(\frac{8}{2^n}=2\)

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

\(2^{3-n}=2^1\)

=> 3 -n = 1

n = 2

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

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

=> 2n -1 = 3

2n = 4

n = 2

Answer 2

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

\(\left(-2\right)^n=-8\)Mà \(-8=2^{-3}\)

\(\Rightarrow x=-3\)

b) \(\frac{8}{2^n}=2\Leftrightarrow2^n=\frac{8}{2}\)

\(2^n=4\)  Mà \(4=2^2\Rightarrow x=2\)

c) \(\left(\frac{1}{2}\right)^{2n-1}=\frac{1}{8}\Rightarrow\left(\frac{1}{2}\right)^{2n}:\frac{1}{2}=\frac{1}{8}\)

\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{8}\cdot\frac{1}{2}\)

\(\left(\frac{1}{2}\right)^{2n}=\frac{1}{16}\Leftrightarrow\frac{1}{2^{2n}}=\frac{1}{16}\)   mà\(16=2^4\)

\(2n=4\Rightarrow n=2\)

Vậy .........................

Answer 3

(-2)^3 và 2^2

câu c ko biết làm nhá

Answer 4

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

<=> (-2)ⁿ = -8

<=> n = 3

b. \(\frac{8}{2^n}\)= 2

<=> 2ⁿ = 4

<=> n = 2

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

<=> \(\left(\frac{1}{2}\right)^{2n-1}=\left(\frac{1}{2}\right)^3\)

<=> 2n - 1 = 3

<=> n = 2