若x+y+z=7,x的平方+y的平方+z的平方=11,xyz=6，则x分之1＋y分之1＋z分之1的值

显示全部楼层 · 2009-3-11 20:11:45

首先，x+y+z = 7所以 1/x+y+z = 1/7x^2 + y^2 + z^2 = (x + y + z)^2 - 2(xy + yz + xz)= 49 - 2(xy + yz + xz)= 11所以可以求得 xy + yz + xz = 191/x + 1/y + 1/z = xy + yz + xz / xyz = 19/6所以1/x + 1/y + 1/z = 19/6或者3又1/6也行...

千问 · 2009-3-11 20:11:45

(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)=7=11+2(xy+yz+xz)所以（xy+yz+xz）=21/x+1/y+1/z+(xy+yz+xz)/(xyz)=2/6=1/3...

千问 · 2009-3-11 20:11:45

(x+y+z)^2 = x^2+y^2+z^2+2xy+2xz+2yz = 49 = 11 +2(xy+yz+xz)->xy+yz+xz = 191/x + 1/y +1/z = (yz+xz+xy)/xyz = 19/6...

千问 · 2009-3-11 20:11:45

(x+y+z)^2=x^2+2xy+y^2+2xz+2yz+z^2=49x^2+y^2+z^2=112xy+2yz+2xz=49-11=38xy+xz+yz=191/x+1/y+1/z=xy+xz+yz/xyz=19/6...