Câu hỏi: Cho số phức thỏa . Tính .
A. .
B. .
C. .
D. .
A.
B.
C.
D.
Ta có:
$\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+{{b}^{2}}-12\sqrt{{{a}^{2}}+{{b}^{2}}}=13 \\
& 2b=-10 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+25-12\sqrt{{{a}^{2}}+25}=13 \\
& b=-5 \\
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& \sqrt{{{a}^{2}}+25}=13 \\
& \sqrt{{{a}^{2}}+25}=-1\left( VN \right) \\
\end{aligned} \right. \\
& b=-5 \\
\end{aligned} \right. \Leftrightarrow \left\{ \begin{aligned}
& a=\pm 12 \\
& b=-5 \\
\end{aligned} \right. \Rightarrow \left\{ \begin{aligned}
& a=12 \\
& b=-5 \\
\end{aligned} \right. a>0 S=a+b=7$.
& {{a}^{2}}+{{b}^{2}}-12\sqrt{{{a}^{2}}+{{b}^{2}}}=13 \\
& 2b=-10 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& {{a}^{2}}+25-12\sqrt{{{a}^{2}}+25}=13 \\
& b=-5 \\
\end{aligned} \right.
& \left[ \begin{aligned}
& \sqrt{{{a}^{2}}+25}=13 \\
& \sqrt{{{a}^{2}}+25}=-1\left( VN \right) \\
\end{aligned} \right. \\
& b=-5 \\
\end{aligned} \right.
& a=\pm 12 \\
& b=-5 \\
\end{aligned} \right.
& a=12 \\
& b=-5 \\
\end{aligned} \right.
Đáp án C.