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Câu hỏi: Cho số phức $z=a+bi,(a,b\in \mathbb{R})$ thỏa mãn $\left( 1+i \right)z-\dfrac{3+4i}{2-i}={{\left( 1-i \right)}^{2}}$. Tính $P=10\text{a}+10b$.
A. $P=-42$
B. $P=20$
C. $P=4$
D. $P=2$
A. $P=-42$
B. $P=20$
C. $P=4$
D. $P=2$
Ta có $\left( 1+i \right)z-\dfrac{3+4i}{2-i}={{\left( 1-i \right)}^{2}}\Leftrightarrow z=\dfrac{{{\left( 1-i \right)}^{2}}+\dfrac{3+4i}{2-i}}{1+i}\Leftrightarrow z=\dfrac{3}{10}-\dfrac{1}{10}i$.
Suy ra $a=\dfrac{3}{10};b=-\dfrac{1}{10}$. Khi đó $P=10\text{a}+10b=10.\dfrac{3}{10}+10\left( -\dfrac{1}{10} \right)=2$.
Suy ra $a=\dfrac{3}{10};b=-\dfrac{1}{10}$. Khi đó $P=10\text{a}+10b=10.\dfrac{3}{10}+10\left( -\dfrac{1}{10} \right)=2$.
Đáp án D.