Cho số phức $z=a+bi(a,b\in ℝ)$thỏa mãn $(1+i)z-\frac{3+4i}{2-i}={(1+i)}_{}^2$. Tính P = 10a + 10b

Ta có:

$\begin{array}{l}(1+i)z-\frac{3+4i}{2-i}\text{Invalid <m:msubsup> element}=\iff (1+i)z-\frac{(3+4i)(2+i)}{5}=-2i\\ \iff (1+i)z-\frac{2+11i}{5}=-2i\iff (1+i)z=-2i+\frac{2+11i}{5}\iff (1+i)z=\frac{2+i}{5}\\ \iff z=\frac{2+i}{5(1+i)}\iff z=\frac{(2+i)(1-i)}{5.2}\iff z=\frac{3}{10}-\frac{1}{10}i\\ \Rightarrow \left\{\begin{array}{l}a=\frac{3}{10}\\ b=-\frac{1}{10}\end{array}\right.\Rightarrow P=10a+10b=2\end{array}$

Chọn: D