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Câu hỏi: Trong không gian $Oxyz$, cho ba điểm $A\left( 0;0;-1 \right)$, $B\left( -1;1;0 \right)$, $C\left( 1;0;1 \right)$. Tìm điểm $M$ sao cho $3M{{A}^{2}}+2M{{B}^{2}}-M{{C}^{2}}$ đạt giá trị nhỏ nhất.
A. $M\left( \dfrac{3}{4};\dfrac{1}{2};-1 \right)$.
B. $M\left( -\dfrac{3}{4};\dfrac{1}{2};2 \right)$.
C. $M\left( -\dfrac{3}{4};\dfrac{3}{2};-1 \right)$.
D. $M\left( -\dfrac{3}{4};\dfrac{1}{2};-1 \right)$.
A. $M\left( \dfrac{3}{4};\dfrac{1}{2};-1 \right)$.
B. $M\left( -\dfrac{3}{4};\dfrac{1}{2};2 \right)$.
C. $M\left( -\dfrac{3}{4};\dfrac{3}{2};-1 \right)$.
D. $M\left( -\dfrac{3}{4};\dfrac{1}{2};-1 \right)$.
Giả sử $M\left( x;y;z \right)\Rightarrow \left\{ \begin{aligned}
& \overrightarrow{AM}=\left( x;y;z+1 \right) \\
& \overrightarrow{BM}=\left( x+1;y-1;z \right) \\
& \overrightarrow{CM}=\left( x-1;y;z-1 \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& A{{M}^{2}}={{x}^{2}}+{{y}^{2}}+{{\left( z+1 \right)}^{2}} \\
& B{{M}^{2}}={{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}+{{z}^{2}} \\
& C{{M}^{2}}={{\left( x-1 \right)}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}} \\
\end{aligned} \right.$
$\Rightarrow 3M{{A}^{2}}+2M{{B}^{2}}-M{{C}^{2}}=3\left[ {{x}^{2}}+{{y}^{2}}+{{\left( z+1 \right)}^{2}} \right]+2\left[ {{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}+{{z}^{2}} \right]$
$-\left[ {{\left( x-1 \right)}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}} \right]$
$=4{{x}^{2}}+4{{y}^{2}}+4{{z}^{2}}+6x-4y+8z+6={{\left( 2x+\dfrac{3}{2} \right)}^{2}}+{{\left( 2y-1 \right)}^{2}}+{{\left( 2z+2 \right)}^{2}}-\dfrac{5}{4}\ge -\dfrac{5}{4}$.
Dấu $''=''$ xảy ra $\Leftrightarrow x=-\dfrac{3}{4}$, $y=\dfrac{1}{2}$, $z=-1$, khi đó $M\left( -\dfrac{3}{4};\dfrac{1}{2};-1 \right)$.
& \overrightarrow{AM}=\left( x;y;z+1 \right) \\
& \overrightarrow{BM}=\left( x+1;y-1;z \right) \\
& \overrightarrow{CM}=\left( x-1;y;z-1 \right) \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& A{{M}^{2}}={{x}^{2}}+{{y}^{2}}+{{\left( z+1 \right)}^{2}} \\
& B{{M}^{2}}={{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}+{{z}^{2}} \\
& C{{M}^{2}}={{\left( x-1 \right)}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}} \\
\end{aligned} \right.$
$\Rightarrow 3M{{A}^{2}}+2M{{B}^{2}}-M{{C}^{2}}=3\left[ {{x}^{2}}+{{y}^{2}}+{{\left( z+1 \right)}^{2}} \right]+2\left[ {{\left( x+1 \right)}^{2}}+{{\left( y-1 \right)}^{2}}+{{z}^{2}} \right]$
$-\left[ {{\left( x-1 \right)}^{2}}+{{y}^{2}}+{{\left( z-1 \right)}^{2}} \right]$
$=4{{x}^{2}}+4{{y}^{2}}+4{{z}^{2}}+6x-4y+8z+6={{\left( 2x+\dfrac{3}{2} \right)}^{2}}+{{\left( 2y-1 \right)}^{2}}+{{\left( 2z+2 \right)}^{2}}-\dfrac{5}{4}\ge -\dfrac{5}{4}$.
Dấu $''=''$ xảy ra $\Leftrightarrow x=-\dfrac{3}{4}$, $y=\dfrac{1}{2}$, $z=-1$, khi đó $M\left( -\dfrac{3}{4};\dfrac{1}{2};-1 \right)$.
Đáp án D.