令x/a=m, y/b=n, z/c=p
m+n+p=1, 1/m+1/n+1/p=0, 求m^2+n^2+p^2的值.
1/m+1/n+1/p=0, mn+np+mp=0
(m+n+p)^2=m^2+n^2+p^2+2(mn+np+mp)=m^2+n^2+p^2=1
所以：m^2+n^2+p^2=1,即所求的值是1.
或者:
x/a+y/b+z/c=1得:bcx+acy+abz=abc
a/x+b/y+c/z=0得:ayz+bxz+cxy=0
由ayz+bxz+cxy=0得:
abc(cxy+bxz+ayz)=0
2abc(cxy+bxz+ayz)=0
2abc^2xy+2ab^2cxz+2a^2bcyz=0
由bcx+acy+abz=abc得:
a^2b^2c^2=a^2b^2z^2+b^2c^2x^2+a^2b^2z^2+2abc^2xy+2ab^2cxz+2a^2bcyz
因为:2abc^2xy+2ab^2cxz+2a^2bcyz=0,所以:
a^2b^2c^2=a^2b^2z^2+b^2c^2x^2+a^2b^2z^2
两边同时除a^2b^2c^2得:
1=z^2/c^2+x^2/a^2+z^2/c^2
即:x^2/a^2+y^2/b^2+z^2/c^2=1
第二题目中是否是:+(1-1/x)^4,?