平方差
原式=(x+y+z-x+y+z)(x+y+z+x-y-z)
=2(y+z)(2x)
=4xy+4xz已知x/3=y/4=z/6≠0，求（xy+yz+xz)/(x^2+y^2+z^2)的值= =令x/3=y/4=z/6=k(k≠0)x=3k,y=4k,z=6k(xy+yz+xz)=12k^2+24k^2+18k^2=54k^2(x^2+y^2+z^2)=9k^2+16k^2+36k^2=61k^2（xy+yz+xz)/(x^2+y^2+z^2)=54k^2/61k^2=54/61