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附加题已知:x-y=a,z-y=10,求x2+y2+z2-xy-yz-zx的最小值.
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附加题
已知:x-y=a,z-y=10,求x2+y2+z2-xy-yz-zx的最小值.
已知:x-y=a,z-y=10,求x2+y2+z2-xy-yz-zx的最小值.
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答案和解析
∵x-y=a,z-y=10,
∴x-a=a-10,
原式=
(2x2+2y2+2z2-2xy-2zx-2yz)
=
[(x-y)2+(y-z)2+(x-z)2]
=
[a2+100+(a-10)2]
=
(2a2-20a+200)
=a2-10a+100
=(a-5)2+75;
所以当a=5时,原式最小值为75
∴x-a=a-10,
原式=
1 |
2 |
=
1 |
2 |
=
1 |
2 |
=
1 |
2 |
=a2-10a+100
=(a-5)2+75;
所以当a=5时,原式最小值为75
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