Question

Решите неравенство (x²-3х-4)(х²-7x+12)≤0

Answer 1

(x²-3x-4)(x²-7x+12)≤0

1) x²-3x-4=0

x₁=-1; x₂=4 (a-b+c=0 [1+3-4=0], x₁=-1; x₂=-c/a=4/1=4)

2) x²-7x+12=0

D=(-7)²-4*12=49-48=1

x₁=(7+1)/2=8/2=4

x₂=(7-1)/2=6/2=3

(x+1)(x-4)(x-3)(x-4)≤0

(x+1)(x-3)(x-4)²≤0

(x-4)²≥0 при x∈R

(x+1)(x-3)≤0

    +       -        +

____.____.____

       -3        1

x∈[-3;1]

Ответ: x∈[-3;1]