(x2+xy) / (x2-y2) = x(x+y) / (x-y)(x+y) = x/ (x-y) = 1 / (x/x -y/x) = 1 / (1 -y/x)
x=2,4,y=0,4
1 / (1 -0,4/2,4) = 1 / (1 -1/6) = 1 / (5/6) = 6/5 = 1,2
ответ 1,2
(a-x)(x³-y³)-(x-y)(a³-x³)=
=(a-x)(x-y)(x²+xy+y²)-(x-y)(a-x)(a²+ax+x²)=
=(a-x)(x-y)(x²+xy+y²-a²-ax-x²)=
=(a-x)(x-y)(y²-a²+xy-ax)
2x³-2xy²-6x²+6y²=
=(2x³-2xy²)-(6x²-6y²)=
=2x(x²-y²)-6(x²-y²)=
=(x²-y²)(2x-6)=
=2(x-3)(x-y)(x+y)
5a²-5b²-10a³b+10ab³=
=(5a²-5b²)-(10a³b-10ab³)=
=5(a²-b²)-10ab(a²-b²)=
=(a²-b²)(5-10ab)=
=5(1-5ab)(a-b)(a+b)
36x³-144x-36x²+144=
=(36x³-36x²)-(144x-144)=
=36x²(x-1)-144(x-1)=
=(x-1)(36x²-144)=
=(x-1)(6x-12)(6x+12)=
=(x-1)*6(x-2)*6(x+2)=
=36(x-1)(x-2)(x+2)
y³+ay²-b²y-b²a=
=(y³+ay²)-(b²y+b²a)=
=y²(y+a)-b²(y+a)=
=(y+a)(y²-b²)=
=(y+a)(y-b)(y+b)
V₁ км/ч - скорость I поезда
V₂ км/ч - скорость II поезда
По условию задачи составим систему уравнений:
{ 10(V₁+ V₂) = 650
{ 4 ¹/₃ V₁ + 8(V₁ + V₂) = 650
{ V₁ + V₂ = 650 : 10
{ 4 ¹/₃ V₁ + 8V₁ + 8V₂ = 650
{ V₁ + V₂ = 65 | * ( - 8)
{ 12 ¹/₃ V₁ + 8V₂ = 650
{ - 8V₁ - 8V₂ = - 520
{12 ¹/₃ V₁ + 8V₂ = 650
Способ сложения:
- 8V₁ - 8V₂ + 12 ¹/₃ V₁ + 8V₂ = - 520 + 650
4 ¹/₃ V₁ = 130
V₁ = 130 : 4 ¹/₃ = ¹³⁰/₁ * ³/₁₃ = 10*3
V₁ = 30 (км/ч) скорость I поезда
30 + V₂ = 65
V₂ = 65 - 30
V₂ = 35 (км/ч) скорость II поезда
Ответ : V₁ = 30 км/ч , V₂ = 35 км/ч .