Question

alternante i,a (2x3+5xi)-3x2=4x5 ax4-15=(80-72)x5 (6x4-ix3)+9x4=3x10+3x3 96-(8x4+ax3)=5x6+19 96-(9x5-9xi)=5x6+57 96-(5x5+ax6-40)x2=9x10. Repede dau coroana!!!

Answer 1

(6+5i)-6=20
6+5i=20+6
6+5i=26
5i=26-6
5i=20
i=20:5
i=4

(24-3i)+36=30+9
(24-3i)+36=39
24-3i=39-36
24-3i=3
3i=24-3
3i=21
i=21:3
i=7

96-(32+3a)=30+19
96-(32+3a) = 49
32+3a=96-49
32+3a=47
3a=47-32
3a=15
a=15:3
a=5

96-(45-9i)=30+57
96-(45-9i)=87
45-9i=95-87
45-9i=8
9i=45-8
9i=36
i=36:9
i=4

96-(25+6a-40)x2=90
96-(6a-15) =90:2
96-(6a-15)=45
6a-15=96-45
6a-15=51
6a=51+15
6a=66
a=66:6
a=11