Bài 1. 

a) \(\left(3+4i\right)+\left(-1+5i\right)=\left(3-1\right)+\left(4i+5i\right)=2+9i\)

b) \(\left(3-4i\right)-\left(1-5i\right)=\left(3-1\right)-\left(4i-5i\right)=2+i\)

c)\(\left(-3+4i\right)+\left(1-4i\right)=\left(-3+1\right)+\left(4i-4i\right)=-2\)

d) \(\left(3-5i\right)-\left(4+i\right)=\left(3-4\right)-\left(5i+i\right)=-1-6i\)

Bài 2. 

a) \(\left(3+4i\right)\left(-1+5i\right)=3.\left(-1\right)+4i.\left(-1\right)+3.5i+4i.5i\)

\(=-3-4i+15i-20=-23+11i\)

b) \(\left(3-5i\right)-\left(4+i\right)=\left(3-4\right)-\left(5i+i\right)=-1-6i\)

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

a)  ( 3 - 4 i ) 2  = 3 2  − 2.3.4i + ( 4 i ) 2  = −7 − 24i

b)  ( 2 + 3 i ) 3  = 2 3  + 3. 2 2 .3i + 3.2. ( 3 i ) 2  + ( 3 i ) 3  = −46 + 9i

c)  [ ( 4   +   5 i )   –   ( 4   + 3 i ) ] 5  = ( 2 i ) 5  = 32i

d)  ( 2 - i 3 ) 2  = −1 − 2i 6

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

b) 2ix + 3 = 5x + 4i

c) 3x(2 – i) + 1 =2ix(1 + i) + 3i

a) 2i(3 + i)(2 + 4i) = 2i(2 + 14i) = -28 + 4i

b)

c) 3 + 2i + (6 + i)(5 + i) = 3 + 2i + 29 + 11i = 32 + 13i

d) 4 - 3i + = 4 - 3i + = 4 - 3i +

= (4 + ) - (3 + )i =

Đáp án là C