(X^2+6x+5<0    Решите системы неравенств. (x^2+4x+3>0 

x²-6x+5<0x²-6x+5=0D=b² - 4ac ⇒ 36 - 4 * 1 * 5 = 36 - 20 = √16 = 4x1 = -b+4/2a ⇒ 6+4/2 = 5x2 = -b-4/2a ⇒ 6-4/2 = 1(x-5)(x-1)Нули: 5, 1-------+---------(1)---------_----------(5)-----+------>x ∈ (1 ; 5)x²+4x+3>0x²+4x+3=0D = 16 - 4 * 1 * 3 = 16 - 12 = √4 = 2x1 = -4 + 2/2 = -1x2 = -4 - 2/2 = -3(x+1)(x+3)Нули: -1, -3-+---(-3)---_----(-1)-------------------------+--------------------->x∈ (-∞ ; -3 ) U (-1 ; +∞)---------------------------(1)--------------------------(5)----------------------->--------(-3)------------(-1)-------------------------------------------------------->Ответ: x ⊂ (-3 ; 5)