问题 解答题
已知x-y=
1+
3
2
z-y=
1-
3
2
,求x2+y2+z2-xy-yz-xz的值.
答案

x-y=

1+
3
2
z-y=
1-
3
2

∴x-z=

3

x2+y2+z2-xy-yz-xz

=

1
2
×(2x2+2y2+2z2-2xy-2yz-2xz)

=

1
2
×[(x-y)2+(y-z)2+(x-z)2]

=

1
2
×[(
1+
3
2
2+(
1-
3
2
2+(
3
2]

=

1
2
×[1+
3
2
+1-
3
2
+3]

=

1
2
×5

=2.5.

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