Ta có: (-2 - 3i) + (-1 - 7i) = (-2 - 1) + (-3 - 7)i = -3 - 10i

(3 + 5i) + (2 + 4i) = (2 + 3) + (5i + 4i) = 5 + 9i.
(-2 - 3i) + (-1 -7i) = (-2 -1) + (-3i -7i) = -3 - 10i.

a) Ta có: (3-2i)(2-3i)=(3.2-2.3)+(-3.3-2.2)i=-13i

b) Ta có: (-1+i)(3+7i)=(-1.3-1.7)+(-1.7+1.3)i=-10-4i

c) Ta có: (5(4+3i)=5.4+5.3i=20+15i

d) Ta có: (-2-5i)4i=(-2.0+5.4)+(2.4-5.0)i=20-8i

\(|4x-3|-|x+7|=0\)

\(\Leftrightarrow|4x-3|=|x+7|\)

\(\Leftrightarrow\orbr{\begin{cases}4x-3=x+7\\4x-3=-x-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x=10\\5x=-4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\\x=\frac{-4}{5}\end{cases}}\)

Vậy \(x\in\left\{\frac{10}{3};\frac{-4}{5}\right\}\)

\(\left|4x-3\right|-\left|x+7\right|=0\)

\(\Leftrightarrow\left|4x-3\right|=\left|x+7\right|\)(1)

* Nếu \(x\ge-7\)thì \(x+7\ge0\Rightarrow\left|x+7\right|=x+7\)

\(\Rightarrow\left(1\right)\Leftrightarrow x+7=\left|4x-3\right|\)(2)

 +) Nếu \(x\ge\frac{4}{3}\)thì \(\left(2\right)\Leftrightarrow x+7=4x-3\Leftrightarrow x=\frac{10}{3}\left(TM\right)\)

 +) Nếu \(x< \frac{4}{3}\)thì \(\left(2\right)\Leftrightarrow x+7=3-4x\Leftrightarrow x=\frac{-4}{5}\left(TM\right)\)

* Nếu \(x< -7\)thì \(x+7< 0\Rightarrow\left|x+7\right|=-x-7\)

\(\Rightarrow\left(1\right)\Leftrightarrow-x-7=\left|4x-3\right|\)(3)

 +) Nếu \(x\ge\frac{4}{3}\)thì \(\left(3\right)\Leftrightarrow-x-7=4x-3\Leftrightarrow x=\frac{-4}{5}\left(TM\right)\)

 +) Nếu \(x< \frac{4}{3}\)thì \(\left(3\right)\Leftrightarrow-x-7=3-4x\Leftrightarrow x=\frac{3}{10}\left(TM\right)\)

Vậy \(x\in\left\{\frac{-4}{5};\frac{3}{10}\right\}\)

Ta có z = z 1 + z 2 = (5-7i) + (2+3i) = 7 - 4i

Chọn A.