Question

1.2.3 помогите
пжжжжж




.
.

ю
​

Answer 1

Ответ: Варiант 2

1. a) (a-4)(a+4) = a^2 - 16

б) (5x+y)(y-5x) = (y+5x)(y-5x) = y^2 - 25

в) (1-x^3)(x^3+1) = (1-x^3)(1+x^3) = 1-x^6

2. a) 49-x^2 = (7-x)(7+x)

б) - 0,01a^2 +b^2 = b2-0,01a^2 = (b-0,1a)(b+0,1a)

в) 9-a^4 = (3-a^2)(3+a^2)

3. a) (x+2)(x-2) - x(x-3)=0

x^2 - 4 - x^2 + 3x =0

-4 + 3x = 0

3x = 4

x= 4/3

б) x^2 - 1 = 0

x^2 = 1 => x1 = 1, x2 = - 1

:3