m

2024 thì bn này ngoài 20 tuổi r trả lời gì nữa :>

<=> 3³ⁿ⁺⁶ : 3²ⁿ⁺⁴ = 3⁴

<=> 3^3n+6-2n-4 = 3⁴

<=> 3ⁿ⁺² = 3⁴

=> n + 2 = 4

<=> n = 2

Vậy n = 2

\(9^{n-2}+9^n-9^2.82=0\)  

\(9^n\left(9^2+1\right)-9^2.\left(9^2+1\right)=0\)

\(\left(9^2+1\right)\left(9^n-9^2\right)=0\)

\(9^n-9^2=0\)

\(9^n=9^2\)

=> n = 2

\(27^n.9^n=9^{27}:81\)

\(3^{3n}.3^{2n}=3^{54}:3^4\)

\(3^{5n}=3^{50}\)

=> 5n = 50

=> n = 10

mk ghi lại đề nha:

27ⁿ : 9ⁿ = 9²⁷ : 81

(27 : 9)ⁿ = 9²⁷ : 9²

\(\Rightarrow\)     3ⁿ  = 9²⁵

\(\Rightarrow\)      3ⁿ = (3²)²⁵

\(\Rightarrow\)      3ⁿ  = 3⁵⁰

Vậy n = 50

\(27^n.9^n=9^{27}:81\Rightarrow3^{3n}:3^{2n}=3^{54}:3^4=3^{50}\)

\(\Rightarrow3^{5n}=3^{50}\Rightarrow5n=50\Rightarrow n=\frac{50}{5}=10\)

Bài 6 :

a) \(\dfrac{625}{5^n}=5\Rightarrow\dfrac{5^4}{5^n}=5\Rightarrow5^{4-n}=5^1\Rightarrow4-n=1\Rightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{27}=-9\Rightarrow\dfrac{\left(-3\right)^n}{\left(-3\right)^3}=\left(-3\right)^2\Rightarrow\left(-3\right)^{n-3}=\left(-3\right)^2\Rightarrow n-3=2\Rightarrow n=5\)

c) \(3^n.2^n=36\Rightarrow\left(2.3\right)^n=6^2\Rightarrow\left(6\right)^n=6^2\Rightarrow n=6\)

d) \(25^{2n}:5^n=125^2\Rightarrow\left(5^2\right)^{2n}:5^n=\left(5^3\right)^2\Rightarrow5^{4n}:5^n=5^6\Rightarrow\Rightarrow5^{3n}=5^6\Rightarrow3n=6\Rightarrow n=3\)

Bài 7 :

a) \(3^x+3^{x+2}=9^{17}+27^{12}\)

\(\Rightarrow3^x\left(1+3^2\right)=\left(3^2\right)^{17}+\left(3^3\right)^{12}\)

\(\Rightarrow10.3^x=3^{34}+3^{36}\)

\(\Rightarrow10.3^x=3^{34}\left(1+3^2\right)=10.3^{34}\)

\(\Rightarrow3^x=3^{34}\Rightarrow x=34\)

b) \(5^{x+1}-5^x=100.25^{29}\Rightarrow5^x\left(5-1\right)=4.5^2.\left(5^2\right)^{29}\)

\(\Rightarrow4.5^x=4.25^{2.29+2}=4.5^{60}\)

\(\Rightarrow5^x=5^{60}\Rightarrow x=60\)

c) Bài C bạn xem lại đề

d) \(\dfrac{3}{2.4^x}+\dfrac{5}{3.4^{x+2}}=\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{10}}\)

\(\Rightarrow\dfrac{3}{2.4^x}-\dfrac{3}{2.4^8}+\dfrac{5}{3.4^{x+2}}-\dfrac{5}{3.4^{10}}=0\)

\(\Rightarrow\dfrac{3}{2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)+\dfrac{5}{3.4^2}\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)=0\)

\(\Rightarrow\left(\dfrac{1}{4^x}-\dfrac{1}{4^8}\right)\left(\dfrac{3}{2}+\dfrac{5}{3.4^2}\right)=0\)

\(\Rightarrow\dfrac{1}{4^x}-\dfrac{1}{4^8}=0\)

\(\Rightarrow\dfrac{4^8-4^x}{4^{x+8}}=0\Rightarrow4^8-4^x=0\left(4^{x+8}>0\right)\Rightarrow4^x=4^8\Rightarrow x=8\)