Question

"спростіть вираз" ("упростить")

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Answer 1

Используем формулы сокращенного умножения

a² - b² = (a - b)(a + b)

a³ - b³ = (a - b)(a² + ab + b²)

a³ + b³ = (a + b)(a² - ab + b²)

1/

(x - 2)(x² + 2x + 4) - (1 + x)(x² - x + 1) = (x³ - 2³) - (x³ + 1) = -9

2/

(x - 3)(x² + 3x + 9) - x(x + 1)(x - 1) = (x³ - 3³) - x(x² - 1²) = x³ - 27 - x³ + x = x - 27

3/

a(a - 3)(a + 3) - (a + 2)(a² - 2a + 4) = a(a² - 3²) - (a³ + 2³) = a³ - 9a - a³ - 8 = -9a - 8

4/ (a² - 1)(a² + 1)(a⁴⁸ + 1)(a¹² + 1)(a²⁴ + 1)(a⁴ - a² + 1)(a⁴ + a² + 1) = a⁹⁶ - 1

по частям 1. первые 2 скобки 2. 3-4-5 скобки 3. 6 и 7 скобки

(a² - 1)(a² + 1) = a⁴ - 1

(a⁴ - a² + 1)(a⁴ + a² + 1) =  (a⁴ + 1 - a² )(a⁴ + 1 + a²) = (a⁴ + 1)² - (a² )² = a⁸ + 2a⁴ + 1 - a⁴ = a⁸ + a⁴ + 1

(a⁴ - 1)(a⁸ + a⁴ + 1) = a¹² - 1

(a¹² - 1)(a⁴⁸ + 1)(a¹² + 1)(a²⁴ + 1) = (a¹² - 1)(a¹² + 1)(a²⁴ + 1)(a⁴⁸ + 1) = (a²⁴ - 1)(a²⁴ + 1)(a⁴⁸ + 1) = (a⁴⁸ - 1)(a⁴⁸ + 1) = a⁹⁶ - 1

Answer 2

ответ: a⁴⁸ - 1.

Объяснение:

$\left(a\:+1\right)\left(a-1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)\left(a^6+1\right)\left(a^{12}+1\right)\left(a^{24}+1\right)=\\\\=\left(a^2-1\right)\left(a^2-a+1\right)\left(a^2+a+1\right)\left(a^6+1\right)\left(a^{12}+1\right)\left(a^{24}+1\right)=\\\\=\left(a^4-a^3+a-1\right)\left(a^2+a+1\right)\left(a^6+1\right)\left(a^{12}+1\right)\left(a^{24}+1\right)=\\\\=(a^6+a^5+a^4-a^5-a^4-a^3+a^3+a^2+a-a^2-a-1)\left(a^6+1\right)\left(a^{12}+1\right)\left(a^{24}+1\right)=$

$=\left(a^6-1\right)\left(a^6+1\right)\left(a^{12}+1\right)\left(a^{24}+1\right)=$

$=\left(a^{12}-1\right)\left(a^{12}+1\right)\left(a^{24}+1\right)=\left(a^{24}-1\right)\left(a^{24}+1\right)=a^{48}-1$