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Câu hỏi: Trong không gian $Oxyz$, cho các điểm $A\left( 0;0;1 \right)$, $B\left( 0;0;4 \right),$ $C\left( 2;2;1 \right),$ $E\left( 4;0;0 \right),$ $F\left( 3;1;\sqrt{6} \right)$. Xét điểm $M$ thay đổi sao cho $MA=\dfrac{1}{2}MB$ và $MA=MC$. Giá trị lớn nhất của $ME+MF$ bằng
A. $4\sqrt{3+\sqrt{3}}$.
B. $4\sqrt{3+\sqrt{6}}$.
C. $4\sqrt{2+\sqrt{2}}$.
D. $4\sqrt{6+\sqrt{6}}$.
A. $4\sqrt{3+\sqrt{3}}$.
B. $4\sqrt{3+\sqrt{6}}$.
C. $4\sqrt{2+\sqrt{2}}$.
D. $4\sqrt{6+\sqrt{6}}$.
Gọi $M\left( x;y;z \right)$. Khi đó giả thiết tương đương với:
$\left\{ \begin{aligned}
& MA=2MA \\
& MA=MC \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}+{{\left( z-4 \right)}^{2}}=4\left( {{x}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}} \right) \\
& {{x}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-1 \right)}^{2}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=4 \\
& x+y-2=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=2-x \\
& {{x}^{2}}+{{\left( 2-x \right)}^{2}}+{{z}^{2}}=4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=2-x \\
& z=\pm \sqrt{4x-2{{x}^{2}}} \\
\end{aligned} \right.$.
Suy ra:
$\begin{aligned}
& ME+MF=\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}}+\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}+{{\left( z-\sqrt{6} \right)}^{2}}} \\
& =\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-8x+16}+\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x-2y-2\sqrt{6}z+16} \\
& =\sqrt{20-8x}+\sqrt{20-6x-2y-2\sqrt{6}z}==\sqrt{20-8x}+\sqrt{20-6x-2\left( 2-x \right)-2\sqrt{6}z} \\
\end{aligned}$
$=\sqrt{20-8x}+\sqrt{16-6x-2\sqrt{6}z}$
$\le g\left( x \right)=\sqrt{20-8x}+\sqrt{16-4x+2\sqrt{6\left( 4x-2{{x}^{2}} \right)}}\le \underset{{}}{\mathop{\underset{\left[ 0;2 \right]}{\mathop{\max }} g\left( x \right)=g\left( 1-\dfrac{\sqrt{3}}{2} \right)}} =4\sqrt{3+\sqrt{3}}$.
$\left\{ \begin{aligned}
& MA=2MA \\
& MA=MC \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}+{{\left( z-4 \right)}^{2}}=4\left( {{x}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}} \right) \\
& {{x}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}}={{\left( x-2 \right)}^{2}}+{{\left( y-2 \right)}^{2}}+{{\left( z-1 \right)}^{2}} \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+{{y}^{2}}+{{z}^{2}}=4 \\
& x+y-2=0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=2-x \\
& {{x}^{2}}+{{\left( 2-x \right)}^{2}}+{{z}^{2}}=4 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& y=2-x \\
& z=\pm \sqrt{4x-2{{x}^{2}}} \\
\end{aligned} \right.$.
Suy ra:
$\begin{aligned}
& ME+MF=\sqrt{{{\left( x-4 \right)}^{2}}+{{y}^{2}}+{{z}^{2}}}+\sqrt{{{\left( x-3 \right)}^{2}}+{{\left( y-1 \right)}^{2}}+{{\left( z-\sqrt{6} \right)}^{2}}} \\
& =\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-8x+16}+\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-6x-2y-2\sqrt{6}z+16} \\
& =\sqrt{20-8x}+\sqrt{20-6x-2y-2\sqrt{6}z}==\sqrt{20-8x}+\sqrt{20-6x-2\left( 2-x \right)-2\sqrt{6}z} \\
\end{aligned}$
$=\sqrt{20-8x}+\sqrt{16-6x-2\sqrt{6}z}$
$\le g\left( x \right)=\sqrt{20-8x}+\sqrt{16-4x+2\sqrt{6\left( 4x-2{{x}^{2}} \right)}}\le \underset{{}}{\mathop{\underset{\left[ 0;2 \right]}{\mathop{\max }} g\left( x \right)=g\left( 1-\dfrac{\sqrt{3}}{2} \right)}} =4\sqrt{3+\sqrt{3}}$.
Đáp án A.