a) (3 + 4i)z = (2 + 5i) – (1 – 3i) = 1 + 8i

Vậy

b) (4 + 7i)z – (5 – 2i) = 6iz ⇔ (4 + 7i)z – 6iz = 5 – 2i

⇔ (4 + i)z = 5 – 2i

a) (1 + 2i)x = − 3 −2i

b) (2 − 2i)x = −(11 + 3i)

x = 12 3 + 8 3 i

(2 − 2i)x = −(11 + 3i)

(1 + 2i)x = − 3 −2i

a) (3 + 2i)z – (4 + 7i) = 2 – 5i

<=> z = \(\dfrac{\text{6 + 2 i}}{\text{3 + 2 i}}\) = \(\dfrac{22}{13}\) - \(\dfrac{6}{13}\)i

b) (7 – 3i)z + (2 + 3i) = (5 – 4i)z

⇔z= \(\dfrac{\text{− 2 − 3 i}}{\text{2 + i}}\) = \(\dfrac{-7}{5}\) - \(\dfrac{4}{5}i\)

c) z² – 2z + 13 = 0

⇔ (z – 1)² = -12 ⇔ z = 1 ± 2 √3 i

d) z⁴ – z² – 6 = 0

⇔ (z² – 3)(z² + 2) = 0

⇔ z ∈ { √3, - √3, √2i, - √2i}

a) đặc : \(x=a+bi\) với \(a;b\in R\) và \(i^2=-1\)

ta có : \(\left(1+2i\right)x-4\left(4-5i\right)=-7+3i\)

\(\Leftrightarrow\left(1+2i\right)\left(a+bi\right)-4\left(4-5i\right)=-7+3i\)

\(\Leftrightarrow a-2b+2ai+bi-16+20i=-7+3i\)

\(\Leftrightarrow\left(a-2b-16\right)+\left(2a+b+20\right)i=-7+3i\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-2b-16=-7\\2a+b+20=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-2b=9\\2a+b=-17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a=-5\\b=-7\end{matrix}\right.\) vậy \(x=-5-7i\)

câu b lm tương tự nha .