Question

我有很多問題

請問這兩題怎麼解？

京若xyz≠0,且
設
2c
6 3
y
z
x+2y-1 z+3
||
X
3
d 4
,
則
6ac-bd
4bd-12ac
則x:y:z=

Answer 1

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Mer

令原式 = k
x = (x+2)k = kx+2k
x = -2k/(k-1)

y = (y-1)k
y = k/(k-1)

z = (z+3)k
z = -3k/(k-1)
所以x : y : z = -2 : 1 : -3

Mer

3a = 2b
a : b = 2 : 3 = 2x : 3x
3d = 4c
c : d = 3 : 4 = 3y : 4y

(6*2x*3y-3x*4y)/(4*3x*4y-12*2x*3y)
= -1