Question

1.a/(b+c)=b/(a+c)=c/(a+b)=k,求k值.

Answer 1

当a+b+c=0时,

b+c=-a

a+c=-b

a+b=-c

这样k=a/-a=b/-b=c/-c=-1

当a+b+c≠0时

把三式相加有

k=(a+b+c)/(2a+2b+2c)=1/2

所以k=-1或1/2

2.先看一部分.

1+a=1+x/(y+z)=(x+y+z)/(y+z)

a/(a+1)=x/(y+z)÷(x+y+z)/(y+z)

=x/(x+y+z)

同理

b/(b+1)=y/(x+y+z)

c/(c+1)=z/(x+y+z)

所以原式=(x+y+z)/(x+y+z)=1

3.两边同乘以(x-2)(x-5)有

1=A(x-5)+B(x-2)

Ax-5A+Bx-2B-1=0

(A+B)x-(5A+2B+1)=0

根据恒等易知

A+B=0

5A+2B+1=0

解得

A=-1/3,B=1/3