知xyz≠0，xy=12(x＋y )，yz= ( y＋z )， ，xz=2 (x＋z )，求x＋y＋z?

因為 xyz ≠ 0 , 所以 x ≠ 0 , y ≠ 0 , z ≠ 0
xy = 12x + 12y
y(x-12) = 12x
y = 12x / (x-12)

xz = 2x + 2z
z(x-2) = 2x
z = 2x / (x-2)

yz = y + z
[ 12x/(x-12) ] * [ 2x/(x-2) ] = [ 12x/(x-12) ] + [ 2x/(x-2) ]
12x * 2x = 12x(x-2) + 2x(x-12)
24x^2 = 12x^2 - 24x + 2x^2 -24x
10x^2 = - 48x
10x = - 48 , 因為 x ≠ 0
x = - 48/10 = - 24/5 = - 4.8

y = 12x/(x-12) = (-57.6)/(-16.8) = 24/7
z = 2x/(x-2) = (-9.6)/(-6.8) = 24/17

x+y+z
= - 24/5 + 24/7 + 24/17
= 24( -1/5 + 1/7 + 1/17 )
= 24( -7*17 + 5*17 + 5*7 )/(5*7*17)
= 24/595

Ans: 24/595

xy=12(x+z)
xy=12x+12y
y(x-12)=12x
y=(12x)/(x-12) i
yz=y+z ii
x+z=2(x+z)
x+z=2x+2z
z=-x iii
代i及iii入ii
(12x)(-x)/(x-12)=-x+[(12x)/(x-12)]
(12x^2)/(12-x)=-x-(12x/12-x)
12x^2=-12x+(x^2)-12x
(11x^2)+24x=0
x(11x+24)=0
x=0 (捨去) 或 x=24/11
代x=24/11入i
y=(12*24/11)/(24/11 -12)
y=-8/3
代x=24/11入iii
z=-24/11
所以(x,y,z)=(24/11,-8/3,-24/11)
x+y+z
=(24/11)-(8/3)-(24/11)
=-8/3