Question

1) 12a^2 - 12b^2

2) 81x^2 - 16x^4

3) (4y - 3)^2 - (2y + 5)^2 = 0

СРОЧНО

Answer 1

$1)12a {}^{2} - 12 {b}^{2} = 12(a + b)(a - b) \\ 2)81x {}^{2} - 16x {} ^{4} = {x}^{2} (9 + 4x)(9 - 4x)$
$3)(4y - 3) {}^{2} - (2y + 5) { }^{2} = 0 \\ (4y - 3 + 2y + 5)(4y - 3 - 2y - 5) = 0 \\ (6y + 2)(2y - 8) = 0 \\ (3y + 1)(y - 4) = 0 \\ y = - \frac{1}{3} \: or \: y = 4$