Question

Запишите в виде мн-на.: (4а^2-3с)^2 и (2а^2+3c)(4а^2-6аc+9c^2); Разложите на множители: 16а^4-25а^2 и -4х^4+8х^2-4; (21^2-34^2) :(56^2-59^2)

Answer 1

1)

(4a² - 3c)² = 16a⁴ - 24a²c + 9c²

2)

(2a²+3c)(4a² - 6ac + 9c²) = 8a⁴ - 12a³c + 18a²c² + 12a²c - 18ac² + 27c³

3)

16a⁴ - 25a² = a²(4a - 5)(4a + 5) = (4a² - 5a)(4a² + 5a)

4)

-4x⁴ + 8x² - 4 = -4(x²-1)² = -4((x-1)(x+1))²

5)

$(21^2-34^2) :(56^2-59^2)=\\\\=(21-34)(21+34):(56-59)(56+59)=\\\\=(21-34)(21+34)(\frac{1}{(56-59)(56+59)} )$

Answer 2

1. Запись в виде многочлена:

a) (4a^2 - 3c)^2 = (4a^2 - 3c)(4a^2 - 3c) [Using the formula for squaring a binomial: (a - b)^2 = a^2 - 2ab + b^2] = 4a^2 * 4a^2 + 4a^2 * (-3c) + (-3c) * 4a^2 + (-3c) * (-3c) [Expanding the expression] = 16a^4 - 24a^2c + 9c^2 [Combining like terms]

b) (2a^2 + 3c)(4a^2 - 6ac + 9c^2) = 2a^2 * 4a^2 + 2a^2 * (-6ac) + 2a^2 * 9c^2 + 3c * 4a^2 + 3c * (-6ac) + 3c * 9c^2 [Expanding the expression] = 8a^4 - 12a^3c + 18a^2c^2 + 12a^2c - 18ac^3 + 27c^3 [Combining like terms]

2. Разложение на множители:

a) 16a^4 - 25a^2 = a^2(16a^2 - 25) [Taking out common factor a^2] = a^2(4a + 5)(4a - 5) [Recognizing this as a difference of squares]

b) -4x^4 + 8x^2 - 4 = -4(x^4 - 2x^2 + 1) [Taking out common factor -4] = -4((x^2)^2 - 2(x^2) + 1) [Recognizing this as a difference of squares] = -4(x^2 - 1)^2 [Factored as a difference of squares]

3. Выражение (21^2 - 34^2) : (56^2 - 59^2) рассчитаем:

(21^2 - 34^2) = (441 - 1156) = -715 (56^2 - 59^2) = (3136 - 3481) = -345

Теперь поделим -715 на -345:

(-715) / (-345) = 715 / 345 ≈ 2.07246377 (округлено до 8 знаков после запятой)

Итак, (21^2 - 34^2) : (56^2 - 59^2) ≈ 2.07246377.