Решить пример: a в 4 степени минус 16. заранее огромное : )

  а^{4} -16=( а^{2}-4)(  а^{2}+4)=(а-2)(а+2)(а^{2}+4) 

x1 = -re(acos(-3)) + 2*pi - i*im(acos(-3))

x2 = 2*pi - i*im(acos(4))

x3 = re(acos(-3)) + i*im(acos(-3))

x4 = re(acos(4)) + i*im(acos(4))

Объяснение:

x1 = -re(acos(-3)) + 2*pi - i*im(acos(-3))

x2 = 2*pi - i*im(acos(4))

x3 = re(acos(-3)) + i*im(acos(-3))

x4 = re(acos(4)) + i*im(acos(4))

x1 = 3.14159265358979 + 1.76274717403909*i

x2 = 6.28318530717959 - 2.06343706889556*i

x3 = 3.14159265358979 - 1.76274717403909*i

x4 = 2.06343706889556*i

сумма

-re(acos(-3)) + 2*pi - i*im(acos(-3)) + 2*pi - i*im(acos(4)) + i*im(acos(-3)) + re(acos(-3)) + i*im(acos(4)) + re(acos(4))

=

4*pi + re(acos(4))

произведение

(((-re(acos(-3)) + 2*pi - i*im(acos(-3)))*(2*pi - i*im(acos(4*(i*im(acos(-3)) + re(acos(-3*(i*im(acos(4)) + re(acos(4)))

=

-(2*pi - i*im(acos(4)))*(i*im(acos(-3)) + re(acos(-3)))*(i*im(acos(4)) + re(acos(4)))*(-2*pi + i*im(acos(-3)) + re(acos(-3)))

Ax²+bx+c=0,d=b²-4ac, x1=(-b+√d)/2a, x2=(-b-√d)/2a 1)5x²+4x-1=0, d=16+20=36,√d=6     x1=(-4+6)/10=2/10=1/5,x1=1/5     x2=(-4-6)/10=-10/10=-1, x2=-1 2)3x²+10x+7=0,d=100-84=16,√16=4     x1=(-10+4)/6=-6/6=-1, x1=-1     x2=(-10-4)/6=-14/6=-7/3, x2=-7/3 3)16x²-2x-5=0,d=4+320=324,√324=18     x1=(2+18)/32=20/32=5/8, x1=5/8   x2=(2-18)/32=-16/32=-1/2, x2=-1/2 4)-7x²-4x+11=0     7x²+4x-11=0,d=16+308=324,√324=18     x1=(-4+18)/14=14/14=1,x1=1     x2=(-4-18)/14=-22/14=-11/7,x2=-11/7 5)28x²-36x+11=0,d=1296-1232=64,√64=8   x1=(36+8)/56=44/56=11/14,x1=11/14   x2=(36-8)/56=28/56=1/2, x21/2 6)-23x²-22x+1=0,     23x²+22x-1=0,d=484+92=576,√576=24   x1=(-22+24)/46=2/46=1/23, x1=1/23   x2=(-22-24)/46=-46/46=-1,x2=-1 7)-49x²+21x-2=0     49x²-21x+2=0,d=441-392=49,√49=7     x1=(21+7)/98=28/98=2/7, x1=2/7     x2=(21-7)/98=14/98=1/7, x2=1/7 8)3x²-14x+16=0,d=196-192=4,√4=2   x1=(14+2)/6=16/6=8/3, x1=8/3   x2=(14-2)/6=12/6=2, x2=2