Đáp án đúng: A
Giải chi tiết:Đặt \(\left\{ \begin{array}{l}2{\log _{1- x}}\left( { - xy + y - 2x + 2} \right) + {\log _{2 + y}}{\left( {x - 1} \right)^2} = 6\,\,\,\left( 1 \right)\\{\log _{1 - x}}\left( {y + 5} \right) - {\log _{2 + y}}\left( {x + 4} \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{array} \right.\)
Điều kiện \(\left\{ \begin{array}{l}0 < 1 - x \ne 1\\0 < 2 + y \ne 1\\x + 4 > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}- 4 < x < 1,\,\,x \ne 0\\- 2 < y \ne - 1\end{array} \right.\)
\(\begin{array}{l}\left\{ \begin{array}{l}2{\log _{1 - x}}\left( { - xy + y - 2x + 2} \right) + {\log _{2 + y}}{\left( {x - 1} \right)^2} = 6\,\,\,\left( 1 \right)\\{\log _{1 - x}}\left( {y + 5} \right) - {\log _{2 + y}}\left( {x + 4} \right) = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\end{array} \right.\\\left\{ \begin{array}{l}0 < 1 - x \ne 1 \Leftrightarrow 0 \ne x < 1\\0 < 2 + y \ne 1 \Leftrightarrow - 2 < y \ne - 1\end{array} \right.\\\left( 1 \right) \Leftrightarrow 2{\log _{1 - x}}\left[ {\left( {y + 2} \right)\left( {1 - x} \right)} \right] + 2{\log _{2 + y}}\left( {1 - x} \right) = 6\\ \Leftrightarrow 2{\log _{1 - x}}\left( {y + 2} \right) + 2 + 2{\log _{2 + y}}\left( {1 - x} \right) = 6\\ \Leftrightarrow {\log _{1 - x}}\left( {y + 2} \right) + \frac{1}{{{{\log }_{1 - x}}\left( {y + 2} \right)}} = 2\\ \Leftrightarrow \log _{_{1 - x}}^2\left( {y + 2} \right) - 2{\log _{1 - x}}\left( {y + 2} \right) + 1 = 0\\ \Leftrightarrow {\log _{1 - x}}\left( {y + 2} \right) = 1 \Leftrightarrow 1 - x = y + 2\end{array}\)
Thế vào (2) ta được
\(\begin{array}{l}{\log _{1 - x}}\left( {4 - x} \right) - {\log _{1 - x}}\left( {x + 4} \right) = 1\\ \Leftrightarrow {\log _{1 - x}}\frac{{4 - x}}{{x + 4}} = 1\\ \Leftrightarrow 1 - x = \frac{{4 - x}}{{x + 4}}\\ \Leftrightarrow x + 4 - {x^2} - 4x = 4 - x\\ \Leftrightarrow {x^2} + 2x = 0 \Leftrightarrow \left[ \begin{array}{l}x = 0\,\,\,\left( {tm} \right) \Rightarrow y = - 1\,\,\left( {ktm} \right)\\x = - 2\,\,\left( {tm} \right) \Rightarrow y = 1\,\,\left( {tm} \right)\end{array} \right.\end{array}\)
Vậy hệ có nghiệm duy nhất là \(\left( {x;y} \right) = \left( { - 2;1} \right)\).
