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Câu hỏi: Tìm số phức $z$ thỏa mãn $\left( 2-3i \right)z-\left( 9-2i \right)=\left( 1+i \right)z$.
A. $-1-2i$.
B. $1-2i$.
C. $\dfrac{13}{5}+\dfrac{16}{5}i$.
D. $1+2i$.
A. $-1-2i$.
B. $1-2i$.
C. $\dfrac{13}{5}+\dfrac{16}{5}i$.
D. $1+2i$.
Ta có $\left( 2-3i \right)z-\left( 9-2i \right)=\left( 1+i \right)z$ $\Leftrightarrow \left( 2-3i \right)z-\left( 1+i \right)z=9-2i$ $\Leftrightarrow \left( 1-4i \right)z=9-2i$ $\Leftrightarrow z=\dfrac{9-2i}{1-4i}$ $\Leftrightarrow z=\dfrac{\left( 9-2i \right)\left( 1+4i \right)}{\left( 1-4i \right)\left( 1+4i \right)}$ $\Leftrightarrow z=\dfrac{17+34i}{17}$ $\Leftrightarrow z=1+2i$
Đáp án D.