Вопрос:
Задание 66. Разложите многочлен на множители с помощью формул сокращенного умножения:
Ответ:
Решение:
- \( a^2 - 4 = (a-2)(a+2) \)
- \( x^2 + 2x + 1 = (x+1)^2 \)
- \( 4x^2 - 4x + 1 = (2x-1)^2 \)
- \( 25 - c^2 = (5-c)(5+c) \)
- \( 4y^2 - 49 = (2y-7)(2y+7) \)
- \( b^2 - 6b + 9 = (b-3)^2 \)
- \( x^2 - 1 = (x-1)(x+1) \)
- \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \)
- \( x^2 - 8x + 16 = (x-4)^2 \)
- \( 8x^3 - y^3 = (2x-y)(4x^2 + 2xy + y^2) \)
- \( \frac{1}{9}a^2 - \frac{2}{9}ab + \frac{1}{9}b^2 = \frac{1}{9}(a^2 - 2ab + b^2) = \frac{1}{9}(a-b)^2 \)
- \( 0.25p^2 - 25t^2 = (0.5p-5t)(0.5p+5t) \)
- \( h^2m^2 - 36c^2 = (hm-6c)(hm+6c) \)
- \( h^3 - 27x^3 = (h-3x)(h^2 + 3hx + 9x^2) \)
- \( 9 + 25c^2 - 30c = 25c^2 - 30c + 9 = (5c-3)^2 \)
- \( n^3 - 27p^6 = (n-3p^2)(n^2 + 3np^2 + 9p^4) \)
- \( 121 + 16y^2 - 88y = 16y^2 - 88y + 121 = (4y-11)^2 \)
- \( 4a^2 + \frac{1}{4}b^2 + 2ab = (2a + \frac{1}{2}b)^2 \)
- \( -8k^3 + 125t^3 = 125t^3 - 8k^3 = (5t-2k)(25t^2 + 10tk + 4k^2) \)
- \( -36x^2 + \frac{1}{49}c^2 = \frac{1}{49}c^2 - 36x^2 = (\frac{1}{7}c - 6x)(\frac{1}{7}c + 6x) \)