(2 − 2i)x = −(11 + 3i)

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 .

(1 + 2i)x = − 3 −2i

Lời giải:

a) 

$(3-2i)x+(5-7i)y=1-3i$

$\Leftrightarrow (3x+5y)-(2x+7y)i=1-3i$
\(\Leftrightarrow \left\{\begin{matrix} 3x+5y=1\\ 2x+7y=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{-8}{11}\\ y=\frac{7}{11}\end{matrix}\right.\)

b) 

\((1+2i)^2x-(4-5i)y=2i\)

\(\Leftrightarrow (-3+4i)x-(4-5i)y=2i\)

\(\Leftrightarrow -(3x+4y)+(4x+5y)i=2i\Leftrightarrow \left\{\begin{matrix} -(3x+4y)=2\\ 4x+5y=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x=10\\ y=-8\end{matrix}\right.\)

(3 + 2i).z - (4 + 7i) = 2 – 5i

⇔ (3 + 2i).z = (2 – 5i) + (4 + 7i)

⇔ (3 + 2i).z = 6 + 2i

(3 - 2i)z + (4 + 5i) = 7 + 3i

⇔ (3 – 2i).z = (7 + 3i) – (4 + 5i)

⇔ (3 – 2i).z = 3 – 2i

⇔ z = 1.

(3 + 4i)x = (1 + 2i)(4 + i)

a) Ta có (3 - 2i)z + (4 + 5i) = 7 + 3i <=> (3 - 2i)z = 7 + 3i - 4 - 5i

<=> z = <=> z = 1. Vậy z = 1.

b) Ta có (1 + 3i)z - (2 + 5i) = (2 + i)z <=> (1 + 3i)z -(2 + i)z = (2 + 5i)

<=> (1 + 3i - 2 - i)z = 2 + 5i <=> (-1 + 2i)z = 2 + 5i

z =

Vậy z =

c) Ta có + (2 - 3i) = 5 - 2i <=> = 5 - 2i - 2 + 3i

<=> z = (3 + i)(4 - 3i) <=> z = 12 + 3 + (-9 + 4)i <=> z = 15 -5i