Question

РЕШИТЕ ПЖ
(x+2)(x+4)^2(x+6)≤-3


x^2-x-8+12/x^-x≥0

Answer 1

$1.\ (x+2)(x+4)^2(x+6) leq -3 \ (x^2+8x+12)(x^2+8x+16) leq -3 \ x^4+8x^3+16x^2+8x^3+64x^2+128x+12x^2+96x+192+3 leq 0 \ x^4+16x^3+92x^2+224x+195 leq 0 \ x^4+3x^3+13x^3+39x^2+53x^2+159x+65x+195 leq 0 \ x^3(x+3)+13x^2(x+3)+53x(x+3)+65(x+3) leq 0 \ (x+3)(x^3+13x^2+53x+65) leq 0 \ (x+3)(x^3+5x^2+8x^2+40x+13x+65) leq 0 \ (x+3)(x^2(x+5)+8x(x+5)+13(x+5)) leq 0 \ (x+3)(x+5)(x^2+8x+13) leq 0 \ \ (x+3)(x+5)(x^2+8x+13)=0 \ \ 1) \ x+3=0 \ x=-3 \ \ 2) \ x+5=0 \ x=-5$$\ \ 3) \ x^2+8x+13=0 \ dfrac{D}{4}=16-13= 3 \ x_1= -4+ sqrt{3} \ x_2=-4- sqrt{3}$

__+___-4-√3___-__-5___+___-3__-___-4+√3__+___

Ответ: x∈[-4-√3;-5]U[-3;-4+√3]

решение второго номера в прикрепе