Đáp án đúng: C
Giải chi tiết:Điều kiện: \(\left\{ \begin{array}{l}{\left( {x + 1} \right)^2} > 0\\4 + x > 0\\4 - x > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}x \ne - 1\\ - 4 < x < 4\end{array} \right..\)
\(\begin{array}{l}{\log _4}{\left( {x + 1} \right)^2} + 2 = {\log _{\sqrt 2 }}\sqrt {4 - x} + {\log _8}{\left( {4 + x} \right)^3}\\ \Leftrightarrow {\log _{{2^2}}}{\left( {x + 1} \right)^2} + 2{\log _2}2 = {\log _{{2^{\frac{1}{2}}}}}{\left( {4 - x} \right)^{\frac{1}{2}}} + {\log _{{2^3}}}{\left( {4 + x} \right)^3}\\ \Leftrightarrow \dfrac{1}{2}{\log _2}{\left( {x + 1} \right)^2} + 2{\log _2}2 = 2{\log _2}{\left( {4 - x} \right)^{\frac{1}{2}}} + \dfrac{1}{3}{\log _2}{\left( {4 + x} \right)^3}\\ \Leftrightarrow {\log _2}\left| {x + 1} \right| + {\log _2}4 = {\log _2}\left( {4 - x} \right) + {\log _2}\left( {4 + x} \right)\\ \Leftrightarrow {\log _2}\left( {4.\left| {x + 1} \right|} \right) = {\log _2}\left[ {\left( {4 - x} \right).\left( {4 + x} \right)} \right]\\ \Leftrightarrow 4.\left| {x + 1} \right| = \left( {4 - x} \right).\left( {4 + x} \right)\\ \Leftrightarrow 4.\left| {x + 1} \right| = {4^2} - {x^2} \Leftrightarrow \left| {x + 1} \right| = \dfrac{{{4^2} - {x^2}}}{4}\\ \Leftrightarrow \left[ \begin{array}{l}x + 1 = \dfrac{{{4^2} - {x^2}}}{4}\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge - 1} \right)\\x + 1 = - \left( {\dfrac{{{4^2} - {x^2}}}{4}} \right)\,\,\,\left( {x < - 1} \right)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}4x + 4 = 16 - {x^2}\,\,\,\,\,\,\left( {x \ge - 1} \right)\\4x + 4 = - 16 + {x^2}\,\,\left( {x < - 1} \right)\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}{x^2} + 4x - 12 = 0\,\,\,\,\,\left( {x \ge - 1} \right)\\{x^2} - 4x - 20 = 0\,\,\,\,\left( {x < - 1} \right)\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}\left[ \begin{array}{l}x = 2\,\,\,\,\,\,\,\,\left( {tm} \right)\\x = - 6\,\,\,\,\,\left( {ktm} \right)\end{array} \right.\\\left[ \begin{array}{l}x = 2 + 2\sqrt 6 \,\,\,\left( {ktm} \right)\\x = 2 - 2\sqrt 6 \,\,\,\left( {tm} \right)\end{array} \right.\end{array} \right.\\ \Rightarrow {x_1}^2 + {x_2}^2 = {2^2} + {\left( {2 - 2\sqrt 6 } \right)^2} = 32 - 8\sqrt 6 .\end{array}\)
Chọn C.
