✔ 最佳答案
1.若x-2y+2z-5=0,則√x^2+(y-1)^2+(z+2)^2之最小值為?
Sol
[x^2+(y-1)^2+(z+2)^2]*[1+(-2)^2+2^2]>=[x-2(y-1)+2(z+2)]^2
[x^2+(y-1)^2+(z+2)^2]*9>=(x-2y+2z+6)^2
[x^2+(y-1)^2+(z+2)^2]*9>=11^2
x^2+(y-1)^2+(z+2)^2>=121/9
√x^2+(y-1)^2+(z+2)^2>=11/3
2.設x,y,z為實數,求 (x+2y-2z)/√(x^2+y^2+z^2) 的最大值為?
Sol
x=y=z=0 =>最大值=∞
xyz<>0
(x^2+y^2+z^2)*[1^2+2^2+(-2)^2]>=(x+2y-2z)^2
(x^2+y^2+z^2)*9>=(x+2y-2z)^2
9>=(x+2y-2z)^2/(x^2+y^2+z^2)
3>=(x+2y-2z)/√(x^2+y^2+z^2)>=-3
最大值=3