Log6(x+1)+log6(2x+1) "меньше или равно"1 .

$log_{6}(x+1)+log_{6}(2x+1) \leq 1$
$log_{6}((x+1)*(2x+1)) \leq log_{6}6$

ОДЗ:

0} \atop {2x+1>0}} \right." alt=" \left \{ {{x+1>0} \atop {2x+1>0}} \right." align="absmiddle" class="latex-formula">

-1} \atop {x>-0,5}} \right." alt="\left \{ {{x>-1} \atop {x>-0,5}} \right." align="absmiddle" class="latex-formula">

-0.5" alt="x>-0.5" align="absmiddle" class="latex-formula">

$(x+1)*(2x+1) \leq 6$
$2x^{2}+x+2x+1-6 \leq 0$
$2x^{2}+3x-5 \leq 0$
$2x^{2}+3x-5 = 0, D=9+4*5*2=49$
$x_{1}= \frac{-3-7}{2}=-5$
$x_{2}= \frac{-3+7}{2}=2$

$-5 \leq x \leq 2$

-0.5" alt="x>-0.5" align="absmiddle" class="latex-formula">

Решение: $-0.5 <x \leq 2$