Question

已知x-y=a,z-y=10,求x^2+y^2+z^2-xy-yx-zx的最小值

Answer 1

解：由于x-y=a,z-y=10
得
x-z=a-10
并且由
x²+y²+z²-xy-yx-zx=1/2[(x-y)²+(y-z)²+(z-x)²]=[a²+10²+(10-a)²]/2
=(2a²-20a+200)/2=a²-10a+100=(a-5)²+75≥75。
即最小值为75。

Answer 2

x-y=a
z-y=10
两个式子相减：x-z=a-10
x^2+y^2+z^2-xy-yz-zx
=0.5(x^2+y^2-2xy)+0.5(x^2+z^2-2zx)+0.5(y^2+z^2-2yz)
=0.5[(x-y)^2+(x-z)^2+(y-z)^2]
=0.5[a^2+(a-10)^2+100]
=a^2-10a+100
=(a-5)^2+75
所以最小值是75

Answer 3

=1/2(x-y)^2+(x-z)^2+(y-z)^2=1/2(a^2+(10-a)^2+10^2)
=a^2+100-10a=(a-5)^2+75>=75

答案为75

Answer 4

这个老师教过。
2（x^2+y^2+z^2-xy-xz-zy)=(x-y)^2+(x-z)^2+(z-y)^2=a^2+(a-10)^2+10^2=最后结果你自己算去吧，那不不懂可以问我