Решите уравнения : 1) (x + 2)^2 - (х2 + 2^2) – 2^3 = 0;

2) (3 + х)^2 - (х^2 + 3^2) – 3^2 = 0;

3) x^2 - (x – 7)(x + 7) = 5 - 2(-2 - x);​

Объяснение:

Пусть x^2 + 2x = t, тогда

t^2 - 2t - 3 = 0

D = 4 + 4*3 = 4 + 12 = 16 = 4^2

t1 =( 2 + 4)/2 = 6/2 = 3 ;

t2 = ( 2 - 4)/2 = - 2/2 = - 1

Имеем 2 уравнения

#1

x^2 + 2x = 3

x^2 + 2x - 3 = 0

D = 4 + 4*3 = 4 + 12 = 16 = 4^2

x1 = ( - 2 + 4)/2 = 2/2 = 1;

x2 = ( - 2 - 4)/2 = - 3

#2

x^2 + 2x + 1 = 0

(x + 1)^2 = 0

x = - 1

- 3; - 1; 1