\(\lim\dfrac{\left(-3\right)^n-4.5^{n+1}}{2.4^n+3.5^n}=\lim\dfrac{\left(-3\right)^n+20.5^n}{2.4^n+3.5^n}=\lim\dfrac{\left(-\dfrac{3}{5}\right)^n+20}{2\left(\dfrac{4}{5}\right)^n+3}=\dfrac{0+20}{0+3}=\dfrac{20}{3}\)

\(\lim\dfrac{2^n-3^n+4.5^{n+2}}{2^{n+1}+3^{n+2}+5^{n+1}}=\lim\dfrac{2^n-3^n+100.5^n}{2.2^n+9.3^n+5.5^n}=\lim\dfrac{\left(\dfrac{2}{5}\right)^n-\left(\dfrac{3}{5}\right)^n+100}{2\left(\dfrac{2}{5}\right)^n+9\left(\dfrac{3}{5}\right)^n+5}=\dfrac{100}{5}=20\)

\(5^{n+1}-4.5^n=5^n.\left(5-4\right)=5^n.1=5^n\)

Chúc bạn hcọ tốt!!!

a) \(\dfrac{-8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)

\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)

c) \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)

\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)

a) \(-\dfrac{8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)

\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)

c) \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)

\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)

a) \(-\dfrac{8}{31}=\dfrac{-8\cdot101}{31\cdot101}=\dfrac{-808}{3131}\)

\(\dfrac{-789}{3131}=\dfrac{-789}{3131}\)

b) \(\dfrac{11}{2^3\cdot3^4\cdot4^5}=\dfrac{11\cdot2^3\cdot5^3}{2^6\cdot3^4\cdot4^5\cdot5^3}=\dfrac{11000}{2^6\cdot3^4\cdot4^5\cdot5^3}\)

\(\dfrac{29}{2^2\cdot2^4\cdot5^3}=\dfrac{29\cdot3^4\cdot4^5}{2^6\cdot3^4\cdot4^5\cdot5^3}=\dfrac{2405376}{2^6\cdot3^4\cdot4^5\cdot5^3}\)

c) \(\dfrac{1}{n}=\dfrac{n+1}{n\left(n+1\right)}\)

\(\dfrac{1}{n+1}=\dfrac{n}{n\left(n+1\right)}\)

5ⁿ⁺¹-4.5ⁿ

=5n.5-4.5n

=5ⁿ

\(5^{n+1}-4.5^n=5^n.5-4.5^n=5^n.\left(5-4\right)=5^n.1=5^n\)

Bài làm :

\(5^{n+1}-4.5^n\)

\(=5^n.5-4.5^n\)

\(=5^n.\left(5-4\right)\)

\(=5^n.1\)

\(=5^n\)

Học tốt

Ta có: \(5^{n+1}-4.5^n=5^n.5-4.5^n=\left(5-4\right)5^n=5^n\)

CHÚC BẠN HỌC TỐT.........

a) 3x\(^n\) (6x\(^{n-3}\)+1) - 2x\(^n\) ( 9x\(^{n-3}\) - 1)

= 18x\(^{n-2}\) + 3x\(^n\) - 18x\(^{n-2}\) + 2x\(^n\)

= 5x\(^n\)

b) 5\(^{n+1}\) - 4.5\(^n\)

= 5\(^n\) . ( 5-4) = 5\(^n\)