Log 3 ( x 2 +4x−4 x 3 −1)+1 / log 3 (1− x 2 − x 3 +x)− loglog 3 ( x 2 +4x−4 x 3 −1)+1 / log 3 (1− x 2 − x 3 +x)− log 3 (x+1) ≥2.

. ОДЗ: 2х+1>0       x>-0,5              (-0,5;0)υ(0;+∞)            2x+1≠1       x≠0            3x+1>0       x>-1/3            3x+1≠1       x≠02. Пусть log2x+1 (3x+1)=t1/t=1+2t(2t2 +t-1)/t=0   t≠0D=9t=-1          t=1/23. log(2x+1)(3x+1)=-1                        log(2x+1)(3x+1)=1/23x+1=1/(2x+1)                            3x+1=√(2x+1)(3x+1)(2x+1)=1                          9x2 +6x+1-2x-1=06x2 +5x=0                                  9x2 +4x=0x(6x+5)=0                                  x(9x+4)=0x=0   x=-5/6                              x=0     x=-4/9ответ:-4/9