Đáp án:

$S = \left\{\dfrac{-11 -\sqrt{2481}}{10};\dfrac{\sqrt{2481}-11}{10}\right\}$

Giải thích các bước giải:

$\begin{array}{l}\quad \dfrac52\log_5(x+2)^2 - 5 = \log_5(4-x)^5 + \dfrac12\log_5(x+6)^{10}\qquad (*)\\ ĐKXĐ:x < 4;\, x \ne -2;\, x \ne -6\\ (*) \Leftrightarrow \dfrac52\cdot 2\log_5|x+2| - 5\log_5(4-x) - \dfrac12\cdot 10\log_5|x+6| = 5\\ \Leftrightarrow 5\log_5(x+2) - 5\log_5(4-x)-5\log_5(x+6) = 5\\ \Leftrightarrow \log_5\dfrac{x+2}{(4-x)(x+6)} = 1\\ \Leftrightarrow \dfrac{x+2}{(4-x)(x+6)} = 5\\ \Leftrightarrow x+2 = 5(4-x)(x+6)\\ \Leftrightarrow 5x^2 + 11x -118 = 0\\ \Leftrightarrow \left[\begin{array}{l} x = \dfrac{-11 -\sqrt{2481}}{10}\\x = \dfrac{\sqrt{2481}-11}{10}\end{array}\right.\quad (nhận)\\ Vậy\,\,S = \left\{\dfrac{-11 -\sqrt{2481}}{10};\dfrac{\sqrt{2481}-11}{10}\right\}\end{array}$

`D = (-infty; 4)`

`5/(2)log_{5} (x + 2)^{2} - 5 = log_{5} (4 - x)^{5} + 1/(2)log_{5} (x + 6)^{10}`

`-> 5/(2).2.log_{5} (x + 2) - 5log_{5} (4 - x) - 1/(2).10.log_{5} (x + 6) = 5`

`-> 5log_{5} (x + 2) - 5log_{5} (4 - x) - 5log_{5} (x + 6) = 5`

`-> log_{5} (x + 2) - log_{5} (4 - x) - log_{5} (x + 6) = 1`

`-> log_{5} ((x + 2)/((4 - x)(x + 6))) = 1`

`-> (x + 2)/((4 - x)(x + 6)) = 5`

`-> x + 2 = 5(4x + 24 - x^2 - 6x)`

`-> x + 2 = 20x + 120 - 5x^2 - 30x`

`-> 5x^2 + 11x - 118 = 0`

`->` \(\left[ \begin{array}{l}x = \dfrac{\sqrt{2481} - 11}{10}\\x = \dfrac{-\sqrt{2481} - 11}{10}\end{array} \right.\) 

`-> S = {(-\sqrt{2481} - 11)/(10); (\sqrt{2481} - 11)/(10)}`