Question

18^n+3/3^2n+5×2^n-2​

Answer 1

Ответ:

$96$

Пошаговое объяснение:

$\frac{18^{n+3}}{3^{2n+5}\cdot 2^{n-2}}= \frac{(9\cdot 2)^{n+3}}{3^{2n+5}\cdot 2^{n-2}}= \frac{9^{n+3}\cdot 2^{n+3}}{3^{2n+5}\cdot 2^{n-2}}= \frac{(3^2)^{n+3}\cdot 2^{n+3}}{3^{2n+5}\cdot 2^{n-2}}= \frac{3^{2(n+3)}\cdot 2^{n+3}}{3^{2n+5}\cdot 2^{n-2}}= \\ \\= \frac{3^{2n+6}\cdot 2^{n+3}}{3^{2n+5}\cdot 2^{n-2}}= 3^{2n+6-(2n+5)}\cdot 2^{n+3-(n-2)}=3^{2n+6-2n-5}\cdot 2^{n+3-n+2}= \\ \\= 3^1 \cdot 2^5= 3\cdot 32=96$

Answer 2

It seems like you have an expression involving exponents and multiplication. Let's break it down step by step:

Expression: 18^n + 3 / (3^(2n+5) × 2^(n-2))

First, we should deal with the division inside the parentheses:

1. 3 / (3^(2n+5)) can be simplified by subtracting the exponents since both bases are the same:

   3 / 3^(2n+5) = 3^(1 - (2n+5)) = 3^(-2n - 4)

Now, let's rewrite the expression with this simplified division:

18^n + 3^(-2n - 4) × 2^(n-2)

Next, if you want to combine these terms, you can keep them in a single fraction if that's your goal:

(18^n + 3^(-2n - 4) × 2^(n-2)) / 1

Now, if you want to simplify further, you can try to get a common denominator for the terms. To do that, let's express 18 as a combination of 3 and 2:

18 = 3^2 × 2

So, we can rewrite 18^n as (3^2)^n × 2^n, and 2^(n-2) as 2^n / 2^2:

((3^2)^n × 2^n + 3^(-2n - 4) × (2^n / 2^2)) / 1

Now, apply the rules of exponents:

(3^(2n) × 2^n + 3^(-2n - 4) × (2^n / 4)) / 1

Now, you can simplify this expression further, but it's a bit complex, and the simplification depends on the specific values of 'n' that you're working with. If you have a specific value of 'n' in mind, I can help you calculate the result more precisely.