Question

(2 –1)(2 + 1)(22+ 1)(24+ 1)(28+ 1) –216. б) 3(22+ 1)(24+ 1)(28+ 1)(216+ 1) –232.

Answer 1

Воспользуемся формулой разности квадратов, (2 –1)(2 + 1)(2^2+ 1)(2^4+ 1)(2^8+ 1)=(2^2 –1)(2^2+ 1)(2^4+ 1)(2^8+ 1)=(2^4 –1)(2^4+1)(2^8+ 1)=(2^8-1)(2^8+1)=2^16-1, да вычтем 2^16, получим -1

Пункт б аналогичен, достаточно представить 3 в виде 2^2-1 или (2-1)(2+1), чтобы уж совсем одинаково получилось, ответ опять же будет -1

Answer 2

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Answer 3

Let's solve the given expressions step by step:

a) (2 – 1)(2 + 1)(22 + 1)(24 + 1)(28 + 1) – 216

Step 1: Perform the operations inside the parentheses first. (2 – 1) = 1 (2 + 1) = 3 (22 + 1) = 3 (24 + 1) = 25 (28 + 1) = 29

Step 2: Multiply all the results together. 1 * 3 * 3 * 25 * 29 = 65,925

Step 3: Subtract 216. 65,925 - 216 = 65,709

Therefore, the result of the expression is 65,709.

b) 3(22 + 1)(24 + 1)(28 + 1)(216 + 1) – 232

Step 1: Perform the operations inside the parentheses first. (22 + 1) = 23 (24 + 1) = 25 (28 + 1) = 29 (216 + 1) = 217

Step 2: Multiply all the results together. 3 * 23 * 25 * 29 * 217 = 6,159,255

Step 3: Subtract 232. 6,159,255 - 232 = 6,159,023

Therefore, the result of the expression is 6,159,023.