(-3; 2; -5); (3; -2; 5)
Объяснение:
(x+y)(x+z)=8
(x+y)(y+z)=3
(y+z)(x+z)=24
(x+y)(x+z)(x+y)(y+z)(y+z)(x+z)=8·3·24
(x+y)²(x+z)²(y+z)²=24·24
[(x+y)(x+z)(y+z)]²=24²
(x+y)(x+z)(y+z)=±24
1) (x+y)(x+z)(y+z)=-24
-24=(x+y)(x+z)(y+z)=8(y+z)⇒y+z=-3
-24=(x+y)(x+z)(y+z)=3(x+z)⇒x+z=-8
-24=(x+y)(x+z)(y+z)=24(x+y)⇒x+y=-1
y+z=-3
x+z=-8
x+y=-1
(y+z)+(x+z)+(x+y)=-3+(-8)+(-1)
2(x+y+z)=-12
x+y+z=-6
x=(x+y+z)-(y+z)=-6-(-3)=-3
y=(x+y+z)-(x+z)=-6-(-8)=2
z=(x+y+z)-(x+y)=-6-(-1)=-5
2) (x+y)(x+z)(y+z)=24
24=(x+y)(x+z)(y+z)=8(y+z)⇒y+z=3
24=(x+y)(x+z)(y+z)=3(x+z)⇒x+z=8
24=(x+y)(x+z)(y+z)=24(x+y)⇒x+y=1
y+z=3
x+z=8
x+y=1
(y+z)+(x+z)+(x+y)=3+8+1
2(x+y+z)=12
x+y+z=6
x=(x+y+z)-(y+z)=6-3=3
y=(x+y+z)-(x+z)=6-8=-2
z=(x+y+z)-(x+y)=6-1=5
1. Область определения
y = 2cos(x-п/3)
D(y) = R
2. Область значения
- 1 ≤ 2cos(x-п/3) ≤ 1
- 1/2 ≤ cos(x-п/3) ≤ 1/2
1) cos(x-п/3) ≥ - 1/2
- arccos(-1/2) + 2πk ≤ x - п/3 ≤ arccos(-1/2) + 2πk, k ∈ Z
- 2π/3 + 2πk ≤ x - п/3 ≤ 2π/3 + 2πk, k ∈ Z
- 2π/3 + π/3 + 2πk ≤ x ≤ 2π/3 + π/3 + 2πk, k ∈ Z
- π/3 + 2πk ≤ x ≤ π + 2πk, k ∈ Z
2) cos(x-п/3) ≤ - 1/2
arccos(-1/2) + 2πk ≤ x - п/3 ≤ 2π - arccos(-1/2) + 2πk, k ∈ Z
2π/3 + 2πk ≤ x - п/3 ≤ 2π - 2π/3 + 2πk, k ∈ Z
2π/3 + 2πk ≤ x - п/3 ≤ 4π/3 + 2πk, k ∈ Z
2π/3 + π/3 + 2πk ≤ x ≤ 4π/3 + π/3 + 2πk, k ∈ Z
π + 2πk ≤ x ≤ 5π/3 + 2πk, k ∈ Z