Question

3．若$\frac{y+z-x}{x+y+z}$=$\frac{z+x-y}{y+z-x}$=$\frac{x+y-z}{z+x-y}$=p，则p+p²+p³的值为1．

Answer 1

分析 根据$\frac{y+z-x}{x+y+z}$=$\frac{z+x-y}{y+z-x}$=$\frac{x+y-z}{z+x-y}$=p，三式相乘得到$\frac{x+y-z}{x+y+z}$=p³，两式相乘得到$\frac{z+x-y}{x+y+z}$=p²，再把$\frac{y+z-x}{x+y+z}$=p代入p+p²+p³求值即可．

解答 解：∵$\frac{y+z-x}{x+y+z}$=$\frac{z+x-y}{y+z-x}$=$\frac{x+y-z}{z+x-y}$=p，
∴$\frac{（y+z-x）（z+x-y）（x+y-z）}{（x+y+z）（y+z-x）（z+x-y）}$=p³，
$\frac{x+y-z}{x+y+z}$=p³，
$\frac{（y+z-x）（z+x-y）}{（x+y+z）（y+z-x）}$=p²，
$\frac{z+x-y}{x+y+z}$=p²，
∴p+p²+p³
=$\frac{y+z-x}{x+y+z}$+$\frac{z+x-y}{x+y+z}$+$\frac{x+y-z}{x+y+z}$
=$\frac{y+z-x+z+x-y+x+y-z}{x+y+z}$
=$\frac{x+y+z}{x+y+z}$
=1．
故答案为：1．

点评 考查了比例的性质，关键是得到p+p²+p³=$\frac{y+z-x}{x+y+z}$+$\frac{z+x-y}{x+y+z}$+$\frac{x+y-z}{x+y+z}$．