Решите уравнения (x+2)^4-4(x+2)^2-5=0

( Х + 2 )^2 = А ; A > 0  ----------------- A^2 - 4A - 5 = 0 D = 16 + 20 = 36 ; √ D = 6  A1 = ( 4 + 6) : 2 = 5 ( > 0 ) A2 = ( 4 - 6) : 2 = ( - 1 ) ( < 0 ) ----------------------- ( X + 2)^2 = 5  X^2 + 4X + 4 - 5 = 0  X^2 + 4X - 1 = 0  D = 16 + 4 = 20  √ D = √ 20 = 2 √ 5  X1 = ( - 4 + 2 √ 5 ) : 2 = ( - 2 + √ 5 )  X2 = ( - 4 - 2 √ 5 ) : 2 = ( - 2 - √ 5 )