Đáp án:
\(\left[ \begin{array}{l}
x = 8\sqrt 5 - 14\\
x = 16 - 8\sqrt 5
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
{x^2} - 2x + 8 - 4\sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} = 0\\
Dat\,\,t = \sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} \,\,\left( {t \ge 0} \right)\\
\Rightarrow {t^2} = \left( {4 - x} \right)\left( {x + 2} \right)\\
\,\,\,\,\,\,{t^2} = 4x + 8 - {x^2} - 2x\\
\,\,\,\,\,\,{t^2} = - {x^2} + 2x + 8\\
\Rightarrow {x^2} - 2x = 8 - {t^2}\\
\Rightarrow Pt:\,\,8 - {t^2} + 8 - 4t = 0\\
\Leftrightarrow - {t^2} - 4t + 16 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t = - 2 + 2\sqrt 5 \,\,\left( {tm} \right)\\
t = - 2 - 2\sqrt 5 \,\,\left( {ktm} \right)
\end{array} \right.\\
\Rightarrow - 2 + 2\sqrt 5 \, = \sqrt {\left( {4 - x} \right)\left( {x + 2} \right)} \\
\Rightarrow 24 - 8\sqrt 5 = - {x^2} + 2x + 8\\
\Leftrightarrow {x^2} - 2x + 16 - 8\sqrt 5 = 0\\
\Delta ' = 1 - 16 + 8\sqrt 5 = 8\sqrt 5 - 15\\
\Rightarrow \left[ \begin{array}{l}
x = 1 + 8\sqrt 5 - 15 = 8\sqrt 5 - 14\\
x = 1 - 8\sqrt 5 + 15 = 16 - 8\sqrt 5
\end{array} \right.
\end{array}\)
