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以知x/a-b=y/b-c=z/c-a(a,b,c互不相等),求x+y+z的值
设x/(a-b)=y/(b-c)=z/(c-a)=k,则x=k(a-b),y=k(b-c),z=k(c-a),于是x+y+z=k(a-b+b-c+c-a)=k*0=0
仿照上述方法解答下列问题
已知:(y+z)/x=(z+x)/y=(x+y)/z,且x+y+z不等于0,求(x+y-z)/(x+y+z)的值

(y+z)/x = (z+x)/y = (x+y)/z(y+z)/x+1 = (z+x)/y+1 = (x+y)/z+1(y+z+x)/x = (z+x+y)/y = (x+y+z)/z = kx+y+z = kx = ky = kz若k=0, 则x+y+z=0,所求表达式无意义若k≠0, 则x=y=z(x+y-z)/(x+y+z)= (x+x-x)/(x+x+x)=1/...