Единицу заряда кулон определяли посредством массы серебра, выделяющегося при электролизе раствора соли серебра. Сформулируйте это определение.

а) 4,17^2-(7,42^2+3/14*0,41) =(4,17-7,42)(4,17+7,42)-3/14*0,41=
=-3,25*11,59- 123/1400=-37 267/400-123/1400=-37 1869/2800 -246/2800=
=-37 2115/2800=-37 423/560
б) 5 12/25*2,8-(6,63^2-3,83^2) =5 12/25-(6,63-3,83)(6,63+3,83)=5,48-2,8*10,46=
=5,48-29,288=-23,808
в) (17,31^2-12,69^2)-(29,81^2-0,19^2)=(17,31-12,69)(17,31+12,69)-(29,81-0,19)(29,81+0,19)=4,62*30 - 29,62*30=30(4,62-29,62)=30*(-25)=-750
г) (7,84^2-12,16^2)+(25,66^2-5,66^2)=(7,84-12,16)(7,84+12,16)+(25,66-5,66)(25,66+5,66) =-4,32*20+20*31,32=20(-4,32+31,32)=20*27=540
д) 0,45^3-1,45^3-0,45*1,45=(0,45-1,45)(0,45²+0,45*1,45+1,45²)-0,45*1,45=
=-1* (0,45²+0,45*1,45+1,45²)-0,45*1,45=-0,45²-0,45*1,45-1,45²-0,45*1,45=
=-(0,45²+2*0,45*1,45+1,45²)=-(0,45+1,45)²=-1,9²=-3,61
е) 0,507^3+0,494^3-0,507*0,493=(0,507+0,494)(0,507²-0,507*0,494+0,494²)-
-0,507*0,494=1,001*(0,507²-0,507*0,494+0,494²)-0,507*0,494=
=0,507²-0,507*0,494+0,494² +0,001*(0,507²-0,507*0,494+0,494²)-0,507*0,494=
=0,507²-2*0,507*0,494+0,494² +0,001*(0,507²-0,507*0,494+0,494²)=
=(0,507-0,494)²+0,001*(0,507²-0,507*0,494+0,494²)=
=0,013²+0,001*(0,507²-0,507*0,494+0,494²)=0,013²+0,001*(0,507²-0,494*0,013)=
=0,000169+0,001*(0,257049-0,006422)=0,000169+0,001*0,250627=
=0,000169+0,000250627=0,000419627

n∈Z

Пусть n = 2, n = 3, n = 4, n = 5

x₁ = (-1)² × π/6 - 5π/12 + π×2/2 = 1 × π/6 - 5π/12 + π = π/6 - 5π/12 + π = 2π/12 - 5π/12 + 12π/12 = 9π/12 = 3π/4

x₂ = (-1)³ × π/6 - 5π/12 + π×3/2 = -1 × π/6 - 5π/12 + 3π/2 = -π/6 - 5π/12 + 3π/2 = -2π/12 - 5π/12 + 18π/12 = 11π/12

x₃ = (-1)⁴ × π/6 - 5π/12 + π×4/2 = 1 × π/6 - 5π/12 + 4π/2 = π/6 - 5π/12 + 4π/2 = 2π/12 - 5π/12 + 24π/12 = 21π/12 = 7π/4

x₄ = (-1)⁵ × π/6 - 5π/12 + π×5/2 = -1 × π/6 - 5π/12 + 5π/2 = -π/6 - 5π/12 + 5π/2 = -2π/12 - 5π/12 + 30π/12 = 23π/12

ответ: x₁ = 3π/4, x₂ = 11π/12, x₃ = 7π/4, x₄ = 23π/12