Question

СРОЧНО 7 КЛАСС ДАМ 40 БАЛЛОВ ПОЖАЛУЙСТА

Answer 1

Ответ:

№1

1) (y-2)(y+2) = y²- 4

2) (x-7)(x+7) = x² - 49

3) (3x - 4)(3x + 4) = 9x² - 16

4) (15x - 2c)(2c + 15x) = 450cx² - 60c²x

№2

1) (x²-7)(x²+7) = x² - 49

2) (2y-9)(2y+9) = 4y² - 81

3) (0.2х - 4у)(4у + 0.2х) = 0.04х² - 16у²

4) (1.3х + 2у²)(1.3х - 2у²) = 1.69х² - ((4у)²)²

Answer 2

Ответ:

#1

$(y - 2)(y + 2) = {y}^{2} + 2y - 2y - 4 = {y}^{2} - 4 \\(x - 7)(x + 7) = {x}^{2} + 7x - 7x - 49 = {x}^{2} - 49$

$(3x - 4)(3x + 4) = 9 {x}^{2} + 12x - 12x - 16 = 9 {x}^{2} - 16 \\ (15x - 2c)(2c + 15x) = 30xc + 225 {x}^{2} - 4 {c}^{2} - 30xc = 225 {x}^{2} - 4 {c}^{2}$

#2

$( {x}^{2} - 7)( {x}^{2} + 7) = {x}^{4} + 7 {x}^{2} - 7 {x}^{2} - 49 = {x}^{4} - 49 \\ (2 {y}^{2} - 9)(2 {y}^{2} + 9) = 4 {y}^{4} + 18 {y}^{2} - 18 {y}^{2} - 81 = 4 {y}^{4} - 81$

$(0.2x - 4y)(4y + 0.2x) = 0.8xy + 0.04 {x}^{2} - 16 {y}^{2} - 0.8xy = 0.04 {x}^{2} - 16 {y}^{2} \\ (1.3x + 2 {y}^{2} )(1.3x - 2 {y}^{2}) = 1.69 {x}^{2} - 2.6x {y}^{2} + 2.6x {y}^{2} - 4 {y}^{4} = 1.69 {x}^{2} - 4 {y}^{4}$