Помогите решить: (x+2)^6 < 8

( x + 2 ) ^ 6 < 8 ; ( ( x + 2 ) ^ 6 ) ^ ( 1 / 6 ) < 8 ^ ( 1 / 6 ); ( x + 2 ) ^ ( 6 * 1 / 6 ) < 8 ^ ( 1 / 6 ); ( x + 2 ) ^ ( 1 * 1 / 1 ) < 8 ^ ( 1 / 6 ); ( x + 2 ) ^ 1 < 8 ^ ( 1 / 6 ); x + 2 < 8 ^ ( 1 / 6 ); x + 2 < ( 2 ^ 3 ) ^ ( 1 / 6 ); x + 2 < 2 ^ ( 3 * 1 / 6 ); x + 2 < 2 ^ ( 3 / 6 ); x + 2 < 2 ^ ( 1 / 2 ) ; ( x + 2 ) ^ 2 < ( 2 ^ ( 1 / 2 ) ) ^ 2; ( x + 2 ) ^ 2 < 2 ^ ( 1 / 2 * 2) ; ( x + 2 ) ^ 2 < 2 ^ 1 ; ( x + 2 ) ^ 2 < 2 ; x ^ 2 + 2 * x * 2 + 2 ^ 2 < 2 ; x ^ 2 + 4 * x + 4 < 2 ; x ^ 2 + 4 * x + 4 - 2 < 0 ; x ^ 2 + 4 * x + 2 < 0 ; x ^ 2 + 4 * x + 2 = 0D = b ^ 2 - 4ac = 4 ^ 2 - 4·1·2 = 16 - 8 = 8x1 = ( -4 - √8 ) / ( 2·1) = -2 - √2 ≈ -3.41;x2 = ( -4 + √8 ) / ( 2·1 ) = -2 + √2 ≈ -0.59; Значит, -2 - √2 < x < -2 + √2.