求值2

求值（1）已知x^2+xy+y^

1。 x^2+xy+y^2=0，当x=0时，y=0，则分式无意义，不予考虑 ===> (x/y)^2+(x/y)+1=0 ===> x/y=(-1±√3i)/2 ①当x/y=(-1+√3i)/2时： x/(x+y)=(x/y)/[1+(x/y)]=[(-1+√3i)/2]/[1+(-1+√3i)/2] =(1+√3i)/2 =cos(π/3)+isin(π/3) y/(x+y)=1/[(1/y)+1] =(1-√3i)/2 =cos(-π/3)+isin(-π/3) 所以，[x/(x+y)]^2005+[y/(x+y)]^2005 =[x/(x+y)]*[x/(x+y)]^2004+[y/...全部

  1。 x^2+xy+y^2=0，当x=0时，y=0，则分式无意义，不予考虑 ===> (x/y)^2+(x/y)+1=0 ===> x/y=(-1±√3i)/2 ①当x/y=(-1+√3i)/2时： x/(x+y)=(x/y)/[1+(x/y)]=[(-1+√3i)/2]/[1+(-1+√3i)/2] =(1+√3i)/2 =cos(π/3)+isin(π/3) y/(x+y)=1/[(1/y)+1] =(1-√3i)/2 =cos(-π/3)+isin(-π/3) 所以，[x/(x+y)]^2005+[y/(x+y)]^2005 =[x/(x+y)]*[x/(x+y)]^2004+[y/(x+y))]*[y/(x+y)]^2005 =[x/(x+y)]*[cos(668π)+isin(668π)]+[y/(x+y)]*[cos(-668π)+isin(-668π)] =[x/(x+y)]*(1+0)+[y/(x+y)]*(1+0) =[x/(x+y)]+[y/(x+y)] =1 同理，当x/y=(-1-√3i)/2时也有同样结果。
   2。
   令a/b=b/c=c/a=k 则： a=bk b=ck c=ak 上述等式左右分别相乘得到：abc=abc*k^3 所以，k=1 则，a=b=c 所以，(a+b-c)/(a-b+c)=1。收起