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Câu hỏi: Cho số phức $z=a+bi$ $\left( a, b\in \mathbb{R} \right)$ thỏa mãn $7a+4+2bi=-10+\left( 6-5a \right)i$. Tính $P=\left( a+b \right)\left| z \right|$
A. $P=12\sqrt{17}$.
B. $P=\dfrac{72\sqrt{2}}{49}$.
C. $P=24\sqrt{17}$.
D. $P=\dfrac{-4\sqrt{29}}{7}$.
A. $P=12\sqrt{17}$.
B. $P=\dfrac{72\sqrt{2}}{49}$.
C. $P=24\sqrt{17}$.
D. $P=\dfrac{-4\sqrt{29}}{7}$.
Ta có: $7a+4+2bi=-10+\left( 6-5a \right)i\Leftrightarrow \left\{ \begin{aligned}
& 7a+4=-10 \\
& 2b=6-5a \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-2 \\
& b=8 \\
\end{aligned} \right.$
Suy ra: $P=\left( -2+8 \right)\sqrt{{{\left( -2 \right)}^{2}}+{{8}^{2}}}=6\sqrt{68}=24\sqrt{17}.$
& 7a+4=-10 \\
& 2b=6-5a \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& a=-2 \\
& b=8 \\
\end{aligned} \right.$
Suy ra: $P=\left( -2+8 \right)\sqrt{{{\left( -2 \right)}^{2}}+{{8}^{2}}}=6\sqrt{68}=24\sqrt{17}.$
Đáp án C.