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Câu hỏi: Xét các số phức $z=a+bi$ $\left( a,b\in \mathbb{R} \right)$ thỏa mãn $\left| z-4-3i \right|=\sqrt{5}$. Tính $P=a+b$ khi $\left| z+1-3i \right|+\left| z-1+i \right|$ đạt giá trị lớn nhất.
A. $P=10$
B. $P=4$
C. $P=6$
D. $P=8$
A. $P=10$
B. $P=4$
C. $P=6$
D. $P=8$
Sử dụng BĐT Bunyakovsky
Từ giả thiết $\left| z-4-3i \right|=\sqrt{5}$ $\Leftrightarrow {{\left( a-4 \right)}^{2}}+{{\left( b-3 \right)}^{2}}=5$ $\Leftrightarrow {{a}^{2}}+{{b}^{2}}-8a-6b+20=0$ $\Leftrightarrow {{a}^{2}}+{{b}^{2}}=8a+6b-20$
Mặt khác $T=\left| z+1-3i \right|+\left| z-1+i \right|$ $=\sqrt{{{\left( a+1 \right)}^{2}}+{{\left( b-3 \right)}^{2}}}+\sqrt{{{\left( a-1 \right)}^{2}}+{{\left( b+1 \right)}^{2}}}$
Suy ra ${{T}^{2}}\le \left( {{1}^{2}}+{{1}^{2}} \right)\left[ {{\left( a+1 \right)}^{2}}+{{\left( b-3 \right)}^{2}}+{{\left( a-1 \right)}^{2}}+{{\left( b+1 \right)}^{2}} \right]$ $=2\left[ 2\left( {{a}^{2}}+{{b}^{2}} \right)-4b+12 \right]$ $=2\left[ 2\left( 8a+6b-20 \right)-4b+12 \right]$ $=8\left( 4a+2b-7 \right)$
Dấu = xảy ra khi ${{\left( a+1 \right)}^{2}}+{{\left( b-3 \right)}^{2}}={{\left( a-1 \right)}^{2}}+{{\left( b+1 \right)}^{2}}$ $\Leftrightarrow a-2b=-2$
Lại có $4a+2b=4\left( a-4 \right)+2\left( b-3 \right)+22\le \sqrt{\left( {{4}^{2}}+{{2}^{2}} \right)\left[ {{\left( a-4 \right)}^{2}}+{{\left( b-3 \right)}^{2}} \right]}+22$ $=\sqrt{20.5}+22=32$
Dấu = xảy ra khi $\dfrac{a-4}{4}=\dfrac{b-3}{2}\Leftrightarrow a-2b=-2$
suy ra ${{T}^{2}}\le 8\left( 4a+2b-7 \right)\le 8\left( 32-7 \right)=200$
$\Rightarrow T\le 10\sqrt{2}$
Vậy ${{T}_{\text{max}}}=10\sqrt{2}$ khi $\left\{ \begin{aligned}
& 4a+2b=32 \\
& a-2b=-2 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& a=6 \\
& b=4 \\
\end{aligned} \right. $. Vậy $ a+b=10$.
Từ giả thiết $\left| z-4-3i \right|=\sqrt{5}$ $\Leftrightarrow {{\left( a-4 \right)}^{2}}+{{\left( b-3 \right)}^{2}}=5$ $\Leftrightarrow {{a}^{2}}+{{b}^{2}}-8a-6b+20=0$ $\Leftrightarrow {{a}^{2}}+{{b}^{2}}=8a+6b-20$
Mặt khác $T=\left| z+1-3i \right|+\left| z-1+i \right|$ $=\sqrt{{{\left( a+1 \right)}^{2}}+{{\left( b-3 \right)}^{2}}}+\sqrt{{{\left( a-1 \right)}^{2}}+{{\left( b+1 \right)}^{2}}}$
Suy ra ${{T}^{2}}\le \left( {{1}^{2}}+{{1}^{2}} \right)\left[ {{\left( a+1 \right)}^{2}}+{{\left( b-3 \right)}^{2}}+{{\left( a-1 \right)}^{2}}+{{\left( b+1 \right)}^{2}} \right]$ $=2\left[ 2\left( {{a}^{2}}+{{b}^{2}} \right)-4b+12 \right]$ $=2\left[ 2\left( 8a+6b-20 \right)-4b+12 \right]$ $=8\left( 4a+2b-7 \right)$
Dấu = xảy ra khi ${{\left( a+1 \right)}^{2}}+{{\left( b-3 \right)}^{2}}={{\left( a-1 \right)}^{2}}+{{\left( b+1 \right)}^{2}}$ $\Leftrightarrow a-2b=-2$
Lại có $4a+2b=4\left( a-4 \right)+2\left( b-3 \right)+22\le \sqrt{\left( {{4}^{2}}+{{2}^{2}} \right)\left[ {{\left( a-4 \right)}^{2}}+{{\left( b-3 \right)}^{2}} \right]}+22$ $=\sqrt{20.5}+22=32$
Dấu = xảy ra khi $\dfrac{a-4}{4}=\dfrac{b-3}{2}\Leftrightarrow a-2b=-2$
suy ra ${{T}^{2}}\le 8\left( 4a+2b-7 \right)\le 8\left( 32-7 \right)=200$
$\Rightarrow T\le 10\sqrt{2}$
Vậy ${{T}_{\text{max}}}=10\sqrt{2}$ khi $\left\{ \begin{aligned}
& 4a+2b=32 \\
& a-2b=-2 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& a=6 \\
& b=4 \\
\end{aligned} \right. $. Vậy $ a+b=10$.
Đáp án A.