一.若分式(4x+8)/(x^2-4)为整数,求x的整数值二.若x/2=y/3=z/4,求(x^2+y^2^+z^2)/xy的值三.若1/a-1/b=3,求(2a+3ab-2b)/(a-ab-a)的值四.若k=z/(x+y)=x/(y+z)=y/(x+z),求k8:00之前回答追加50分,由于分式是初学,所以希望大家

Question

一.若分式(4x+8)/(x^2-4)为整数,求x的整数值
二.若x/2=y/3=z/4,求(x^2+y^2^+z^2)/xy的值
三.若1/a-1/b=3,求(2a+3ab-2b)/(a-ab-a)的值
四.若k=z/(x+y)=x/(y+z)=y/(x+z),求k
8:00之前回答追加50分,
由于分式是初学,所以希望大家能每一步都清楚些.

Answer 1

一.(4x+8)/(x^2-4)=4(x+2)/(x+2)(x-2)=4/(x-2)为整数
x=-2,0,1,3,5,6
二.x/2=y/3=z/4,
x/y=2/3,y/x=3/2
z/x=4/2=2,z/y=4/3
(x^2+y^2^+z^2)/xy
=x/y+y/x+z/x*z/y
=2/3+3/2+2*4/3
=29/6
三.1/a-1/b=3---->b-a=3ab
(2a+3ab-2b)/(a-ab-b)
=(2(a-b)+3ab)/((a-b)-ab)
=(-6ab+3ab)/(-3ab-ab)
=-3ab/-4ab
=3/4
四.k=z/(x+y)=x/(y+z)=y/(x+z)
=(z+x+y)/((x+y)+(y+z)+(x+z))
=(z+x+y)/(2(z+x+y))
=1/2