Số phức \(z\) thỏa mãn \(\dfrac{2+i}{1-i}.z=\dfrac{-1+3i}{2+i}\) là
\(\dfrac{22}{25}-\dfrac{4}{25}i\).\(\dfrac{22}{25}+\dfrac{4}{25}i\).\(\dfrac{21}{25}+\dfrac{4}{25}i\).\(\dfrac{21}{25}+\dfrac{4}{25}i\). Hướng dẫn giải: \(\frac{2+i}{1-i}.z=\frac{-1+3i}{2+i}\)
\(\Leftrightarrow z=\frac{-1+3i}{2+i}:\frac{2+i}{1-i}\)
\(\Leftrightarrow z=\frac{\left(-1+3i\right)\left(1-i\right)}{\left(2+i\right)^2}\)
\(\Leftrightarrow z=\frac{2+4i}{3+4i}\)
\(\Leftrightarrow z=\frac{\left(2+4i\right)\left(3-4i\right)}{\left(3+4i\right)\left(3-4i\right)}=\frac{22}{25}+\frac{4}{25}i.\)
