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Câu hỏi: Cho số phức $z=a+bi$ $\left( a,b\in \mathbb{R} \right)$ thỏa mãn $\dfrac{2-iz}{2+i}-\dfrac{z+2i}{1-2i}=2\bar{z}$. Tính $a+b.$
A. 1.
B. 2.
C. −1.
D. −2.
A. 1.
B. 2.
C. −1.
D. −2.
Giả sử $z=a+bi\left( a,b\in \mathbb{R} \right)$
Ta có $\dfrac{2-iz}{2+i}-\dfrac{z+2i}{1-2i}=2\bar{z}\Leftrightarrow \dfrac{\left( 2-iz \right)\left( 2-i \right)}{5}-\dfrac{\left( z+2i \right)\left( 1+2i \right)}{5}=2\left( a-bi \right)$
$\Leftrightarrow \left( 4-2i-2iz-z \right)-\left( z+2iz+2i-4 \right)=10\left( a-bi \right)$
$\Leftrightarrow 8-4i-2z-4iz=10\left( a-bi \right)\Leftrightarrow 8-4i-2\left( a+bi \right)-4i\left( a+bi \right)=10\left( a-bi \right)$
$\Leftrightarrow \left\{ \begin{aligned}
& 8-2\text{a}+4b=10\text{a} \\
& -4-2b-4\text{a}=-10b \\
\end{aligned} \right.\Leftrightarrow a=b=1\Rightarrow a+b=2$.
Ta có $\dfrac{2-iz}{2+i}-\dfrac{z+2i}{1-2i}=2\bar{z}\Leftrightarrow \dfrac{\left( 2-iz \right)\left( 2-i \right)}{5}-\dfrac{\left( z+2i \right)\left( 1+2i \right)}{5}=2\left( a-bi \right)$
$\Leftrightarrow \left( 4-2i-2iz-z \right)-\left( z+2iz+2i-4 \right)=10\left( a-bi \right)$
$\Leftrightarrow 8-4i-2z-4iz=10\left( a-bi \right)\Leftrightarrow 8-4i-2\left( a+bi \right)-4i\left( a+bi \right)=10\left( a-bi \right)$
$\Leftrightarrow \left\{ \begin{aligned}
& 8-2\text{a}+4b=10\text{a} \\
& -4-2b-4\text{a}=-10b \\
\end{aligned} \right.\Leftrightarrow a=b=1\Rightarrow a+b=2$.
Đáp án B.