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Câu hỏi: Cho số phức $z=1+i.$ Biết rằng tồn tại các số phức ${{z}_{1}}=a+5i,{{z}_{2}}=b$ (trong đó $a,b\in \mathbb{R},b>1)$ thoả mãn $\sqrt{3}\left| z-{{z}_{1}} \right|=\sqrt{3}\left| z-{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|.$ Tính $b-a$ thu được kết quả là
A. $b-a=5\sqrt{3}.$
B. $b-a=2\sqrt{3}.$
C. $b-a=4\sqrt{3}.$
D. $b-a=3\sqrt{3}.$
A. $b-a=5\sqrt{3}.$
B. $b-a=2\sqrt{3}.$
C. $b-a=4\sqrt{3}.$
D. $b-a=3\sqrt{3}.$
Ta có $\sqrt{3}\left| z-{{z}_{1}} \right|=\sqrt{3}\left| z-{{z}_{2}} \right|=\left| {{z}_{1}}-{{z}_{2}} \right|\Rightarrow \left\{ \begin{aligned}
& {{\left( 1-a \right)}^{2}}+{{4}^{2}}={{\left( b-1 \right)}^{2}}+1 \\
& {{\left( b-a \right)}^{2}}+25=3\left[ {{\left( 1-a \right)}^{2}}+16 \right] \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}}=15 \\
& {{\left( b-1 \right)}^{2}}+2\left( b-1 \right)\left( 1-a \right)+{{\left( 1-a \right)}^{2}}=3{{\left( 1-a \right)}^{2}}+\dfrac{23}{15}\left[ {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}} \right] \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}}=15 \\
& 8{{\left( b-1 \right)}^{2}}-30\left( b-1 \right)\left( 1-a \right)+7{{\left( 1-a \right)}^{2}}=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}}=15 \\
& \left[ \begin{aligned}
& b-1=\dfrac{1}{4}\left( 1-a \right) \\
& b-1=\dfrac{7}{2}\left( 1-a \right) \\
\end{aligned} \right. \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=1-\dfrac{2\sqrt{3}}{3} \\
& b=\dfrac{7\sqrt{3}}{3}+1 \\
\end{aligned} \right.\Rightarrow b-a=3\sqrt{3}.$
CáCh kháC:
Đặt $\left\{ \begin{aligned}
& u=a-1 \\
& v=b-1 \\
\end{aligned} \right. $ ta có hpt: $ \left\{ \begin{aligned}
& {{v}^{2}}-{{u}^{2}}=15 \\
& {{v}^{2}}+2uv-2{{u}^{2}}=23 \\
\end{aligned} \right.$(Hệ đẳng cấp quen thuộc).
& {{\left( 1-a \right)}^{2}}+{{4}^{2}}={{\left( b-1 \right)}^{2}}+1 \\
& {{\left( b-a \right)}^{2}}+25=3\left[ {{\left( 1-a \right)}^{2}}+16 \right] \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}}=15 \\
& {{\left( b-1 \right)}^{2}}+2\left( b-1 \right)\left( 1-a \right)+{{\left( 1-a \right)}^{2}}=3{{\left( 1-a \right)}^{2}}+\dfrac{23}{15}\left[ {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}} \right] \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}}=15 \\
& 8{{\left( b-1 \right)}^{2}}-30\left( b-1 \right)\left( 1-a \right)+7{{\left( 1-a \right)}^{2}}=0 \\
\end{aligned} \right.$
$\Leftrightarrow \left\{ \begin{aligned}
& {{\left( b-1 \right)}^{2}}-{{\left( 1-a \right)}^{2}}=15 \\
& \left[ \begin{aligned}
& b-1=\dfrac{1}{4}\left( 1-a \right) \\
& b-1=\dfrac{7}{2}\left( 1-a \right) \\
\end{aligned} \right. \\
\end{aligned} \right.\Rightarrow \left\{ \begin{aligned}
& a=1-\dfrac{2\sqrt{3}}{3} \\
& b=\dfrac{7\sqrt{3}}{3}+1 \\
\end{aligned} \right.\Rightarrow b-a=3\sqrt{3}.$
CáCh kháC:
Đặt $\left\{ \begin{aligned}
& u=a-1 \\
& v=b-1 \\
\end{aligned} \right. $ ta có hpt: $ \left\{ \begin{aligned}
& {{v}^{2}}-{{u}^{2}}=15 \\
& {{v}^{2}}+2uv-2{{u}^{2}}=23 \\
\end{aligned} \right.$(Hệ đẳng cấp quen thuộc).
Đáp án D.