Question

Peço que a resposta seja adequadamente detalhada e clara.

Simplifique a expressão $\frac{ 3^{n-3} + 3^{n+1} }{ 3^{n-2} + 3^{n} }$

Answer 1

$\frac{3^{n-3}+3^{n+1}}{3^{n-2}+3^{n}} \\ \\ \frac{3^{n}.3^{-3}+3^{n}.3^1}{3^{n}.3^{-2}+3^{n}} \ \ \ \ \ a^{p+q}=a^p.a^q\\ \\ \frac{3^n(3^{-3}+3)}{3^n(3^{-2}+1)} \ \ \ \ \ a^{-p}= \frac{1}{a^p} \\ \\ \frac{ \frac{1}{3^3}+3 }{ \frac{1}{3^2}+1 } = \frac{ \frac{1}{27}+3 }{ \frac{1}{9}+1 }= \frac{\frac{1+81}{27}}{ \frac{1+9}{9} } = \frac{ \frac{82}{27} }{ \frac{10}{9} }= \frac{82}{27} . \frac{9}{10}= \frac{82}{30}= \frac{41}{15}$

Answer 2

Explicação passo-a-passo:

$\frac{ {3}^{n - 3} + 3 {}^{n + 1} }{ {3}^{n - 2} + {3}^{n} } \\ \\ \frac{( 1 + {3}^{4} ) \times {3}^{n - 3} }{(1 + {3}^{2} ) \times {3}^{n - 2} } \\ \\ \frac{1 + 81}{(1 + 9) \times 3} \\ \\ \frac{82}{10 \times 3} \\ \\ \frac{41}{5 \times 3} \\ \\ \frac{41}{15}$

• Espero ter ajudado.