Question

,нужны оба варианта формулы сокращенного цмножения

Answer 1

Вариант 1

Задание 1

а) $(a^2 - 2b)^2 = a^4 - 4a^2b + 4b^2$

б) $(-7x - 1)^2 = 49x^2 + 14x + 1$

в) $(-5a + 2b^4)^2 = (2b^4-5a)^2= 4b^8 - 20ab^4 + 25a^2$

г) $(x - 12)(x + 12) = x^2 - 144$

д) $(-4n^3 + 7)(4n^3 + 7) = (7 - 4n^3)(7 + 4n^3) - 49 - 16n^6$

Задание 2

а) $18c^2 - 2(3c-1)^2 = 18c^2 - 2(9c^2 - 6c + 1) = 18c^2 - 18c^2 + 12c - 2 = 12c -2 = 2(6c -1)$

б) $(2a + 3)(a -3) - 2a(4 - 6a) = 2a^2 -3a + 3a - 9 - 4a + 12a^2 = 14a^2 - 4a - 9$

в)

$(2x + 3)(3 - 2x) - (2x - 5)^2 - 10x = 9 - 4x^2 - (4x^2- 20x + 25) - 10x = \\\\9 - 4x^2 - 4x^2 + 20x - 25 - 10x = -8x^2 + 10x - 16 = -2(4x^2 - 5x + 8)$

Задание 3

а)

$x(x-2)(x+1) = x^2(x-1)\\\\x(x^2 + x - 2x - 2) = x^3 - x^2\\\\x^3 - x^2 - 2x = x^3 - x^2\\\\x^3 - x^2 - 2x - x^3 + x^2 = 0\\\\-2x = 0\\\\x = 0$

б)

$(2y - 1)^2 - 4(y - 2)(y + 2) = -5y + 18\\\\4y^2 - 4y + 1 - 4y^2 + 16 = -5y + 18\\\\-4y + 5y = 18 - 16 - 1\\\\y = 1$

Вариант 2

Задание 1

а) $(3x - y)^2 = 9x^2 - 6xy + y^2$

б) $(-6n - 2)^2 = 36n^2 + 24n + 4$

в) $(-7c + 8x^3)^2 = (8x^3-7c)^2= 64x^6 - 112cx^3 + 49c^2$

г) $(b - 13a)(b + 13a) = b^2 - 169a^2$

д) $(-5m^4 + 3)(3 + 5m^4) = (3 - 5m^4)(3+5m^4) = 9 - 25m^8$

Задание 2

а)

$-36c - 3(5 - 6c)^2 = - 36c - 3(25 - 50c + 36c^2) = -36c - 75 + 150c - 108c^2 = \\\\-75 + 114c - 108c^2 = -3(25 - 38c + 36c^2)$

б) $(7b - 1)(1 + 7b) - b(4 - 3b) = 49b^2 - 1 - 4b + 3b^2 = 52b^2 - 4b - 1$

в) $(2k - 3)(3 + 2k) - (3k - 5)^2 - 30k = 4k^2 - 9 - 9k^2 + 30k - 25 - 30k = -5k^2 - 34$

Задание 3

а)

$x^2 - (x-4)(x+4) = 2x\\\\x^2 - x^2 + 16 = 2x\\\\16 = 2x\\\\x =8$

б)

$(3y + 1)^2 - 9(y + 1)(y - 1) = 8y - 16\\\\9y^2 + 6y + 1 - 9y^2 + 9 = 8y - 16\\\\6y - 8y = -16 - 10\\\\-2y = -26\\\\y = 13$