Question

log11(8x^2+7) -log11(x^2+x+1)=>log11(x/(x+5) +7)

Answer 1

ОДЗ: x/(x+5) + 7 > 0
(x+7x+35)/(x+5) > 0
(8x+35)/(x+5) > 0
__+___(-5)__-____(-35/8)___+___

$log_{11}(8x^2+7)-log_{11}(x^2+x+1) \geq log_{11}( \frac{x}{x+5} +7)\\\\ log_{11} \frac{8x^2+7}{x^2+x+1} \geq log_{11}( \frac{x}{x+5} +7)\\\\ \frac{8x^2+7}{x^2+x+1} \geq \frac{x}{x+5} +7\\\\ \frac{(8x^2+7)(x+5)-x(x^2+x+1)-7(x^2+x+1)(x+5)}{(x^2+x+1)(x+5)} \geq 0\\\\ \frac{8x^3+40x^2+7x+35-x^3-x^2-x-7x^3-42x^2-42x-35}{(x^2+x+1)(x+5)} \geq 0\\\\ \frac{-3x^2-36x}{(x^2+x+1)(x+5)} \geq 0\\\\ --+-[-12]-----(-5)----+-----[0]-----\\\\ x\in (-\infty;-12]U(-4,375;0]-s-ODZ$

Answer 2

log ₁₁(8x²+7) - log ₁₁(x²+x+1)≥ log ₁₁(x /(x+5) + 7)

ОДЗ   8x²+7> 0 при любом значении х
           x²+x+1 >0 при любом значении х
          х+5≠0  х≠-5

          x /(x+5) + 7  >0 
         
           x+7*(x+5)
         -------------------- >0 
             x+5

           8x+35
         ------------  >0 
             x+5
 $\left \{ {{8x+35\ \textgreater \ 0} \atop x+5\ \textgreater \ 0 }} \right.$

$\left \{ {{x\ \textgreater \ -35/8} \atop {x\ \textgreater \ -5 }} \right.$



 $\left \{ {{8x+35\ \textless \ 0} \atop { x+5\ \textless \ 0}} \right.$

$\left \{ {{x\ \textless \ -35/8} \atop {x\ \textless \ -5}} \right.$

ОДЗ  x∈(-∞;-5)∪(-4 3/8 ;+∞)


log ₁₁(8x²+7) /(x²+x+1)≥ log ₁₁(x /(x+5) + 7)

(8x²+7)/(x²+x+1) - (8x +35) /(x+5)  ≥ 0

(8x²+7)(x+5) -  (8x +35)(x²+x+1)
--------------------------------------------------  ≥ 0
       (x²+x+1)(x+5)

8х³+7х+40х²+35-8х³-35х-8х²-35х²-8х-35
---------------------------------------------------------- ≥ 0
        (x²+x+1)(x+5)

-3х²-36х
-------------------- ≥ 0
 (x²+x+1)(x+5)


    +                  -                +            -
______-12______-5_________0______________

Ответ с учетом ОДЗ  (-∞;-12) ∪ (-4 3/8 ;0]