（初二）化简与求值 可用因式分解法、公式法、设k法、换元法、整体代入法……

Question

63.4
详细过程

Answer 1

a+1=x/(y+x)+1=(x+y+z)/(y+z)
所以1/(a+1)=(y+x)/(x+y+z)
同理1/(b+1)=(z+x)/(x+y+z)
1/(c+1)=(x+y)/(x+y+z)
而所求=(a+1-1)/(a+1)+(b-1+1)/(b+1)+(c-1+1)/(c+1)=3-1/(a+1)-1/(b+1)-1/(c+1)=3-(y+x+z+x+x+y)/(x+y+z)=3-2=1

Answer 2

解：用整体代入法
原式=(x/x+y+z)+(y/x+y+z)+(z/x+y+z)
=(x+y+z)/(x+y+z)
=1