Question

Помогите пожалуйста с 5, 6,7,8,9

Answer 1

a) \((b + 2a)(b - 2a) = b^2 - (2a)^2 = b^2 - 4a^2\)

6) \((8c - 3y)(8c + 3y) = (8c)^2 - (3y)^2 = 64c^2 - 9y^2\)

B) \((b^2 - 3)^2 = b^4 - 6b^2 + 9\)

a) \(7(x + 8) + (x + 8)(x - 8) = 7(x + 8) + (x + 8)(x) - (x + 8)(8)\)

\(= 7x + 56 + x^2 - 8x - 8x - 64\)

\(= x^2 - 9x - 8\)

(x + 5)4x - (2x + 5)2 = \(4x(x + 5) - 2(x + 5) = 4x^2 + 20x - 2x - 10 = 4x^2 + 18x - 10\)

a) \((x + 3)(x - 3) - x(x + 4) = (x^2 - 3^2) - (x^2 + 4x) = x^2 - 9 - x^2 - 4x = -9 - 4x\)

6) \((4x + 1)^2 + (3 - 2x)(8x + 1) = (4x + 1)^2 + (3 \cdot 8x - 2x^2 + 1) = (4x + 1)^2 + (24x - 2x^2 + 1)\)

\(= 16x^2 + 8x + 1 + 24x - 2x^2 + 1 = 14x^2 + 32x + 2\)

8. \((x^2 - 3)(x^2 + 3) - (x^2 - 5)^2 \) при \(x = 3\)

\(= (3^2 - 3)(3^2 + 3) - (3^2 - 5)^2\)

\(= (9 - 3)(9 + 3) - (9 - 5)^2\)

\(= 6 \cdot 12 - 4^2\)

\(= 72 - 16 = 56\)

9. \((3b - 4)^2 - (b + 7)^2 = ((3b - 4) + (b + 7))((3b - 4) - (b + 7))\)

\(= (4b + 3)(2b - 11)\)