Question

18^2017 + 6^4034:3^2015 (12^2017 + 6^2017) Вычислите

Answer 1

Ответ:

9

Пошаговое объяснение:

18²⁰¹⁷+6⁴⁰³⁴÷3²⁰¹⁵(12²⁰¹⁷+6²⁰¹⁷)=6²⁰¹⁷·3²⁰¹⁷+6²⁰¹⁷·6²⁰¹⁷÷3²⁰¹⁵(6²⁰¹⁷·2²⁰¹⁷+6²⁰¹⁷)=(6²⁰¹⁷(3²⁰¹⁷+6²⁰¹⁷))/(3²⁰¹⁵·6²⁰¹⁷(2²⁰¹⁷+1))=(3²⁰¹⁷+6²⁰¹⁷)/(3²⁰¹⁵(2²⁰¹⁷+1))=(3²⁰¹⁷(1+2²⁰¹⁷))/(3²⁰¹⁵(2²⁰¹⁷+1))=3²⁰¹⁷⁻²⁰¹⁵=3²=9

Answer 2

To solve the given expression, you need to follow the order of operations (PEMDAS/BODMAS) and perform calculations step by step. Let's break it down:

1. Calculate the value inside the parentheses first: (12^2017 + 6^2017)

2. Calculate the value outside the parentheses using the value obtained in step 1: 18^2017 + 6^4034:3^2015 * (12^2017 + 6^2017)

Let's start with step 1:

1. Calculate the value inside the parentheses: (12^2017 + 6^2017)

   12^2017 means 12 raised to the power of 2017. 6^2017 means 6 raised to the power of 2017.

To calculate these values, we can use a calculator or a programming language with support for large numbers. After performing the calculations, we can move to step 2:

2. Calculate the value outside the parentheses using the value obtained in step 1: 18^2017 + 6^4034:3^2015 * (12^2017 + 6^2017)

At this point, we have the values for 12^2017 and 6^2017. Now let's continue the calculation:

18^2017 + 6^4034:3^2015 * (12^2017 + 6^2017)

Let's denote the value of 12^2017 as X, and the value of 6^2017 as Y:

X = 12^2017 Y = 6^2017

Now, we can substitute these values back into the expression:

18^2017 + 6^4034:3^2015 * (X + Y)

Remember, to divide by a number raised to a power, you can subtract the exponents:

6^4034:3^2015 = 6^(4034-2015)

Now, calculate this value and continue with the main expression:

6^(4034-2015) = 6^2019

So, the expression becomes:

18^2017 + 6^2019 * (X + Y)

Now, calculate 18^2017, 6^2019, and then add them:

Finally, you'll get the result of the expression. I'm afraid I can't provide the exact numerical result due to the large numbers involved, but you can use a calculator or a programming language with support for big integers to compute it accurately.