Các số thực x, y thỏa mãn \(\frac{{x - 3}}{{3 + i}} + \frac{{y - 3}}{{3 - i}} = i\). Khi đó, tổng T = x+y bằng

\(\begin{array}{*{20}{l}}
{\begin{array}{*{20}{l}}
{\frac{{x - 3}}{{3 + i}} + \frac{{y - 3}}{{3 - i}} = i}\\
{ \Leftrightarrow \frac{{\left( {x - 3} \right)\left( {3 - i} \right)}}{{{3^2} - {i^2}}} + \frac{{\left( {y - 3} \right)\left( {3 + i} \right)}}{{{3^2} - {i^2}}} = \frac{{i\left( {3 - i} \right)\left( {3 + i} \right)}}{{{3^2} - {i^2}}}}
\end{array}}\\
{\begin{array}{*{20}{l}}
{ \Leftrightarrow 3x - xi - 9 + 3i + 3y + yi}\\
{{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu} {\mkern 1mu}  - 9 - 3i = 10i}
\end{array}}\\
{ \Leftrightarrow 3x + 3y - 18 + \left( {y - x} \right)i = 10i}\\
{ \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{ - x + y = 10}\\
{x + y = 6}
\end{array}} \right. \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}
{x =  - 2}\\
{y = 8}
\end{array}} \right.}\\
{ \Rightarrow T = x + y = 6}
\end{array}\)