一道初二代数题
http://bbs.dolchn.com/UploadFile/2007-3/20073142251610936.gif
参考答案:已知xy/(x+y)=a,xz/(x+z)=b,yz/(y+z)=c,且abc≠0,求x的值。
解:将三个已知全部倒过来,得
1/x+1/y=1/a
1/x+1/z=1/b
1/y+1/z=1/c······················①
以上三式相加,得
2(1/x+1/y+1/z)=1/a+1/b+1/c
1/x+1/y+1/z=1/2(1/a+1/b+1/c)
上式减去①式,得
1/x=1/2(1/a+1/b-1/c)
1/x=(bc+ca-ab)/2abc
所以:x=2abc/(bc+ca-ab)。