Найти сумму комплексных чисел 5-2i и 3-3i

1)  (√43-5)²+10(√43-0,2) = 43 - 10√43 + 25 + 10√43 - 2 = 16 2) 14(1+√31)+(7-√31)² = 14+14√31+49-14√31+31= 94 3)  (√2-√15)²+(√6+√5)²= 2 - 2√30 + 15 + 6 + 2√30 + 5 = 28 1)  2√3(√12+3√5)-√5(6√3-√20) = 2√36 + 6√15 - 6√15  +  √100 = 12+10= 22 2)  √6(0,5√24-8√11)-4√11(√99-2√6)=   0,5√144 - 8√66 - 4√1089 + 8√66 = (0,5*12) + (2*33) = 6+66=72 3)  (√162-10√5)√2+(5+√10)² =  √324 - 10√10 + 25 + 10√10 + 10 = 18+25+10=53 4)  (17-√21)²-5√3(4√27-6,8√7)= 289 -  34√21+ 21 - 20√81 + 34√21 = 289 + 21 -  20*9 = 289+21 -  180= 130.

12(√3/2cosx-1/2sinx)=√2 cos(x+π/6)=√2/2 x+π/6=-π/4+2πn u x+π/6=π/4+2πn x=-5π/12+2πn u x=π/12+2πn,n∈z 2 2(1/2cosx-√3/2sinx)=2cos5x cos(x+π/3)=cos5x 5x=x+π/3+2πn u 5x=-π/3-x+2πn 4x=π/3+2πn u 6x=-π/3+2πn x=π/12+πn/2 u x=-π/18+πn/3,n∈z 3 sin3xcos2x=sin(3x+2x) sin3xcos2x=sin3xcos2x+sin2xcos3x sin2xcos3x=0 sin2x=0⇒2x=πn⇒x=πn/2,n∈z cos3x=0⇒3x=π/2+πn⇒x=π/6+πn/3,n∈z 4 sinxsin7x=sin3xsin5x 1/2[cos(7x-x)-cos(7x+x)]=1/2[cos(5x-3x)-cos(5x+3x)] cos6x-cos8x=cos2x-cos8x cos6x=cos2x 6x=2x+2πn u 6x=-2x+2πn 4x=2πn u 8x=2πn x=πn/2 u x=πn/4-общий