0 không thể là ước của bất cứ số nào em nhá

=0

\(\left(\dfrac{3}{7}\right)^2\cdot\left(-7\right)^4\)

\(=\dfrac{3^2}{7^2}\cdot7^4\)

\(=\dfrac{3^2\cdot7^4}{7^2}\)

\(=3^2\cdot7^2\)

\(=\left(3\cdot7\right)^2=441\)

E = -1/3 +1/(3^2) - 1/(3^3) + .... - 1/(3^51)

E.1/3 = -1/(3^2) + 1/(3^3)-1/(3^4) +.... - 1/(3^52)

E + E.1/3 = [-1/3+1/(3^2) - 1/(3^3) +.... -1/(3^51)]+[-1/(3^2) +1/(3^3) -1/(3^4) +.... - 1/(3^52)]

E.4/3 = -1/3-1/(3^52)

E.4/3 = (-3^51 - 1)/(3^52)

E = (-3^51 - 1)/(3^52) . 3/4

E = (-3^51-1)/(4.3^51)

???

a) \(\dfrac{81}{\left(-3\right)^n}=-243\)

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

\(\left(-3\right)^n=\dfrac{\left(-3\right)^4}{\left(-3\right)^5}=\left(-3\right)^{-1}\)

n = -1

Vậy n = -1

b) \(\dfrac{25}{5^n}=5\)

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

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

n = 1

Vậy n = 1

c) \(\dfrac{1}{2}\cdot2^n+4\cdot2^n=9\cdot2^5\)

\(2^{n-1}+4\cdot2^{n-1}\cdot2=9\cdot2^5\)

\(2^{n-1}+8\cdot2^{n-1}=9\cdot2^5\)

\(\left(8+1\right)\cdot2^{n-1}=9\cdot2^5\)

\(9\cdot2^{n-1}=9\cdot2^5\)

\(2^{n-1}=2^5\cdot\dfrac{9}{9}=2^5\)

n - 1 = 5

n = 5 + 1 = 6

Vậy n = 6

a) 81/(-3)ⁿ = -243

(-3)ⁿ = 81 : (-243)

(-3)ⁿ = -1/3

n = -1

b) 25/5ⁿ = 5

5ⁿ = 25 : 5

5ⁿ = 5

n = 1

c) 1/2 . 2ⁿ + 4 . 2ⁿ = 9 . 2⁵

2ⁿ . (1/2 + 4) = 9 . 32

2ⁿ . 9/2 = 288

2ⁿ = 288 : 9/2

2ⁿ = 64

2ⁿ = 2⁶

n = 6