Пожалуйста, решите неравенство (очень прошу прикрепить решение): 1/(3x+3)(3x+4)+ 1/(3x+4)(3x+5)+1/(3x+5)(3x+6)=<1/6

Ответ:

x=(-∞;-3]U(-2;-5/3)U(-5/3;-4/3)U(-4/3;-1)U[0;+∞)

Пошаговое объяснение:

ОДЗ:

   x≠-1

   x≠-(4/3)

   x≠-(5/3)

   x≠-2

1/(3x+3)(3x+4)+ 1/(3x+4)(3x+5)+1/(3x+5)(3x+6)-1/6=<0

1/3(x+1)(3x+4)+ 1/(3x+4)(3x+5)+1/3(3x+5)(x+2)-1/6=<0

(2*(3x+5)*(x+2)+6(x+1)*(x+2)+2*(x+1)*(3x+4)-(x+1)*(3x+4)*(3x+5)*(x+2))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

((6x+10)*(x+2)+(6x+6)*(x+2)+(2x+2)*(3x+4)-(3x²+4x+3x+4)*(3x+5)*(x+2))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(6x²+12x+10x+20+6x²+12x+6x+12+6x²+8x+6x+8-9x-54-119x²-114x-40)/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-101x²-60x-9-54)/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(9x³+54x²+101x+60))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(9x³+27x²+27²+81x+20x+60))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(9x²(x+3)+27x(x+3)+20(x+3)))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(x+3)*(9x²+27x+20))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(x+3)*(9x²+15x+12x+20))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(x+3)*(3x(3x+5)+4(3x+5)))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(x+3)*(3x+5)*(3x+4))/(6*(x+1)*(3x+4)*(3x+5)*(x+2))<=0

(-x*(x+3))/(6*(x+1)*(x+2))<=0

x=(-∞;-3]U(-2;-5/3)U(-5/3;-4/3)U(-4/3;-1)U[0;+∞)