cái j vậy

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a) Ta có: \(3^{54}=\left(3^6\right)^9=729^9\)

Lại có: \(2^{81}=\left(2^9\right)^9=512^9\)

Ta có: \(729^9>512^9\Rightarrow3^{54}>2^{81}\)

b) Ta có: \(5\cdot125\cdot625=5^1\cdot5^3\cdot5^4=5^8\)

Lại có: \(625^3=\left(5^4\right)^3=5^{12}\)

Ta có: \(5^8< 5^{12}\Rightarrow5\cdot125\cdot625< 625^3\)

c) Xét: \(8^4\cdot16^3\cdot32\)

\(=\left(2^3\right)^4\cdot\left(2^4\right)^3\cdot2^5\)

\(=2^{12}\cdot2^{12}\cdot2^5\)

\(=2^{29}\)

Xét: \(64^4\cdot8^2\)

\(=\left(2^6\right)^4\cdot\left(2^3\right)^2\)

\(=2^{24}\cdot2^6\)

\(=2^{30}\)

Ta có: \(2^{29}< 2^{30}\Rightarrow8^4\cdot16^3\cdot32< 64^4\cdot8^2\)

a, 5⁶ : 5ⁿ = 625

=> 5⁶ : 5ⁿ = 5⁴

=> 5ⁿ = 5⁶ : 5⁴

=> 5ⁿ = 5²

=> n = 2

b, n³ = 216

=> n³ = 6³

=> n = 6

c, n² = 2³ + 3² + 4³

=> n² = 8 + 9 + 64

=> n ² = 81

=> n² = 9²

=> n = 9

\(\frac{9^{14}\cdot25^5\cdot8^7}{18^{12}\cdot625^3\cdot24^3}=\frac{3^{28}\cdot5^{10}\cdot2^{21}}{2^{12}\cdot3^{24}\cdot5^{12}\cdot2^9\cdot3^3}=\frac{3\cdot3^{27}\cdot5^{10}\cdot2^{21}}{2^{21}\cdot3^{27}\cdot5^2\cdot5^{10}}=\frac{3}{5^2}=\frac{3}{25}\)

\(\frac{9^{14}.25^5.8^7}{18^{12}.625^3.24^3}\)

\(=\)\(\frac{3^{28}.5^{10}.2^{21}}{2^{12}.3^{24}.5^{12}.2^9.3^3}\)

\(=\)\(\frac{3.3^{27}.5^{10}.2^{21}}{2^{21}.3^{27}.5^2.5^{10}}\)

\(=\)\(\frac{3}{5^2}\)

\(=\)\(\frac{3}{25}\)

\(\dfrac{9^{14}\cdot25^5\cdot8^7}{18^2\cdot625^3\cdot24^3}\)

\(=\dfrac{3^{42}\cdot5^{10}\cdot2^{21}}{2^2\cdot3^4\cdot5^{12}\cdot2^9\cdot3^3}=\dfrac{3^{42}\cdot5^{10}\cdot2^{21}}{2^{11}\cdot3^7\cdot5^{12}}\)

\(=\dfrac{3^{35}}{5^2}\cdot2^{10}\)

a: \(625^5=5^{20}\)

\(125^7=5^{21}\)

mà 20<21

nên \(625^5< 125^7\)

3²ⁿ⁺¹.5=10935 đúng ko

={2n+3}mũ 4=5 mũ 4

=2n+3=5

2n=5-2=3

n=2 chia 2=1

vậy n=1 hok tốtttttt

Trả lời:

3²ⁿ⁺¹.5=10935

=3²ⁿ.3¹.5=10935

=3²ⁿ.3=10935:5=2187

=3²ⁿ=2187:3=729

=3²ⁿ=3⁶

=2n=6

=>2.n=6

=>n=6:2=3

Vậy n=3

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