 
                                                 
                                                Объяснение:
Уравнение к-ой степени имеет к корней в поле комплексных чисел
cos(π/12)=cos[(π/3)-(π/4)]=cos(π/3)cos(π/4)+sin(π/3)sin(π/4)=
=0,5·(√2/2)+(√3/2)(√2/2)=(√6+√2)/4
sin(π/12)=sin[(π/3)-(π/4)]=sin(π/3)cos(π/4)-cos(π/3)sin(π/4)=
=(√3/2)(√2/2)-0,5·(√2/2)=(√6-√2)/4
x⁶+3i=0
x⁶=-3i=3(cos(-π/2)+isin(-π/2))
x₀=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos(-π/12)+isin(-π/12))=
 (cos(-π/12)+isin(-π/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos(π/12)-isin(π/12))=
 (cos(π/12)-isin(π/12))=
=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) ((√6+√2)/4-i(√6-√2)/4)=
((√6+√2)/4-i(√6-√2)/4)=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) ((√6+√2)-i(√6-√2))/4
((√6+√2)-i(√6-√2))/4
x₁=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((-π+2π)/12)+isin((-π+2π)/12))=
 (cos((-π+2π)/12)+isin((-π+2π)/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((π)/12)+isin((π)/12))=
 (cos((π)/12)+isin((π)/12))=
![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) ((√6+√2)+i(√6-√2))/4
((√6+√2)+i(√6-√2))/4
x₂=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((-π+4π)/12)+isin((-π+4π)/12))=
 (cos((-π+4π)/12)+isin((-π+4π)/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((3π)/12)+isin((3π)/12))=
 (cos((3π)/12)+isin((3π)/12))=
=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((π)/4)+isin((π)/4))=
 (cos((π)/4)+isin((π)/4))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (√2/2+i√2/2)=
 (√2/2+i√2/2)=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) √2(1+i)/2
 √2(1+i)/2
x₃=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((-π+6π)/12)+isin((-π+6π)/12))=
 (cos((-π+6π)/12)+isin((-π+6π)/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((5π)/12)+isin((5π)/12))=
 (cos((5π)/12)+isin((5π)/12))=
=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (sin((π)/12)+icos((π)/12))=
 (sin((π)/12)+icos((π)/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) ((√6-√2)/4+i(√6+√2)/4)=
 ((√6-√2)/4+i(√6+√2)/4)=
=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) ((√6-√2)+i(√6+√2))/4
((√6-√2)+i(√6+√2))/4
x₄=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((-π+8π)/12)+isin((-π+8π)/12))=
 (cos((-π+8π)/12)+isin((-π+8π)/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((7π)/12)+isin((7π)/12))=
 (cos((7π)/12)+isin((7π)/12))=
=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (-sin((π)/12)+icos((π)/12))=
 (-sin((π)/12)+icos((π)/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (-(√6-√2)/4+i(√6+√2)/4)=
 (-(√6-√2)/4+i(√6+√2)/4)=
=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) ((√2-√6)+i(√6+√2))/4
((√2-√6)+i(√6+√2))/4
x₅=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((-π+10π)/12)+isin((-π+10π)/12))=
 (cos((-π+10π)/12)+isin((-π+10π)/12))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (cos((3π)/4)+isin((3π)/4))=
 (cos((3π)/4)+isin((3π)/4))=
=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (-cos((π)/4)+isin((π)/4))=
 (-cos((π)/4)+isin((π)/4))=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) (-√2/2+i√2/2)=
 (-√2/2+i√2/2)=![\sqrt[6]{3}](/tpl/images/1081/0740/cb4a8.png) √2(-1+i)/2
 √2(-1+i)/2
 
                                                 
                                                 
                                                 
                                                
Пусть х км/ч - собственная скорость катера, тогда (х + 2) км/ч - скорость катера по течению реки, (х - 2) км/ч - скорость катера против течения реки. На путь туда и обратно катер затратил 9 часов. Уравнение:
80/(х+2) + 80/(х-2) = 9
80 · (х - 2) + 80 · (х + 2) = 9 · (х + 2) · (х - 2)
80х - 160 + 80х + 160 = 9 · (х² - 2²)
160х = 9х² - 36
9х² - 160х - 36 = 0
D = b² - 4ac = (-160)² - 4 · 9 · (-36) = 25600 + 1296 = 26896
√D = √26896 = 164
х₁ = (160-164)/(2·9) = (-4)/18 = - 2/9
х₂ = (160+164)/(2·9) = 324/18 = 18
Вiдповiдь: 18 км/год - власна швидкiсть катера.