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Câu hỏi: Trong không gian với hệ tọa độ Oxyz, cho ba điểm $A\left( a;0;0 \right)$, $B\left( 0;b;0 \right)$, $C\left( 0;0;c \right)$ với a, b, c là những số dương thay đổi thỏa mãn ${{a}^{2}}+4{{b}^{2}}+16{{c}^{2}}=49$. Tính tổng $S={{a}^{2}}+{{b}^{2}}+{{c}^{2}}$ khi khoảng cách từ O đến mặt phẳng $\left( ABC \right)$ đạt giá trị lớn nhất.
A. $S=\dfrac{51}{5}$.
B. $S=\dfrac{49}{4}$.
C. $S=\dfrac{49}{5}$.
D. $S=\dfrac{51}{4}$.
A. $S=\dfrac{51}{5}$.
B. $S=\dfrac{49}{4}$.
C. $S=\dfrac{49}{5}$.
D. $S=\dfrac{51}{4}$.
Phương trình mặt phẳng $\left( ABC \right)$ : $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1\Leftrightarrow \dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}-1=0\left( a,b,c>0 \right)$.
$d\left( O,\left( ABC \right) \right)=\dfrac{\left| \dfrac{0}{a}+\dfrac{0}{b}+\dfrac{0}{c}-1 \right|}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}=\dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}=P$.
${{P}_{\max }}\Leftrightarrow T=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}\min $.
$T=\dfrac{1}{{{a}^{2}}}+\dfrac{4}{4{{b}^{2}}}+\dfrac{16}{16{{c}^{2}}}\ge \dfrac{{{\left( 1+2+4 \right)}^{2}}}{{{a}^{2}}+4{{b}^{2}}+16{{c}^{2}}}=\dfrac{{{7}^{2}}}{49}=1$.
${{T}_{\min }}=1$. Dấu bằng xảy ra khi và chỉ khi
$\left\{ \begin{aligned}
& \dfrac{1}{{{a}^{2}}}=\dfrac{2}{4{{b}^{2}}}=\dfrac{4}{16{{c}^{2}}} \\
& {{a}^{2}}+4{{b}^{2}}+16{{c}^{2}}=49 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{{{a}^{2}}}=\dfrac{2}{4{{b}^{2}}}=\dfrac{4}{16{{c}^{2}}} \\
& {{a}^{2}}+4.\dfrac{{{a}^{2}}}{2}+16.\dfrac{{{a}^{2}}}{4}=49 \\
\end{aligned} \right.\Leftrightarrow {{a}^{2}}=7;{{b}^{2}}=\dfrac{7}{2};{{c}^{2}}=\dfrac{7}{4}$
Vậy $S={{a}^{2}}+{{b}^{2}}+{{c}^{2}}=\dfrac{49}{4}$.
$d\left( O,\left( ABC \right) \right)=\dfrac{\left| \dfrac{0}{a}+\dfrac{0}{b}+\dfrac{0}{c}-1 \right|}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}=\dfrac{1}{\sqrt{\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}}}=P$.
${{P}_{\max }}\Leftrightarrow T=\dfrac{1}{{{a}^{2}}}+\dfrac{1}{{{b}^{2}}}+\dfrac{1}{{{c}^{2}}}\min $.
$T=\dfrac{1}{{{a}^{2}}}+\dfrac{4}{4{{b}^{2}}}+\dfrac{16}{16{{c}^{2}}}\ge \dfrac{{{\left( 1+2+4 \right)}^{2}}}{{{a}^{2}}+4{{b}^{2}}+16{{c}^{2}}}=\dfrac{{{7}^{2}}}{49}=1$.
${{T}_{\min }}=1$. Dấu bằng xảy ra khi và chỉ khi
$\left\{ \begin{aligned}
& \dfrac{1}{{{a}^{2}}}=\dfrac{2}{4{{b}^{2}}}=\dfrac{4}{16{{c}^{2}}} \\
& {{a}^{2}}+4{{b}^{2}}+16{{c}^{2}}=49 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{1}{{{a}^{2}}}=\dfrac{2}{4{{b}^{2}}}=\dfrac{4}{16{{c}^{2}}} \\
& {{a}^{2}}+4.\dfrac{{{a}^{2}}}{2}+16.\dfrac{{{a}^{2}}}{4}=49 \\
\end{aligned} \right.\Leftrightarrow {{a}^{2}}=7;{{b}^{2}}=\dfrac{7}{2};{{c}^{2}}=\dfrac{7}{4}$
Vậy $S={{a}^{2}}+{{b}^{2}}+{{c}^{2}}=\dfrac{49}{4}$.
Đáp án B.