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Câu hỏi: Cho số phức $z=a+bi\text{ (}a,b\in \mathbb{R}\text{)}$ thoả mãn $(1+i)z+2\overline{z}=3+2i$. Tính $P=a+b$
A. $P=1$.
B. $P=-\dfrac{1}{2}$.
C. $P=\dfrac{1}{2}$.
D. $P=-1$
A. $P=1$.
B. $P=-\dfrac{1}{2}$.
C. $P=\dfrac{1}{2}$.
D. $P=-1$
$(1+i)z+2\overline{z}=3+2i\Leftrightarrow (1+i)(a+bi)+2(a-bi)=3+2i\Leftrightarrow (3a-b)+(a-b)i=3+2i$
$\Leftrightarrow \left\{ \begin{aligned}
& 3a-b=3 \\
& a-b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{2} \\
& b=-\dfrac{3}{2} \\
\end{aligned} \right. $. Suy ra: $ P=a+b=-1$.
$\Leftrightarrow \left\{ \begin{aligned}
& 3a-b=3 \\
& a-b=2 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=\dfrac{1}{2} \\
& b=-\dfrac{3}{2} \\
\end{aligned} \right. $. Suy ra: $ P=a+b=-1$.
Đáp án D.