Question

Подсказка, нужно перемножить уравнения
$(x + y)(x + z) = 8 \\ (x + y)(y + z) = 3 \\ (y + z)(x + z) = 24$
Помогите срочно прям ​

Answer 1

Ответ:

(-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