\s\up8(→(→)=a-b,
所以\s\up8(→(→)=\s\up8(→(→)=a-b,
所以\s\up8(→(→)=\s\up8(→(→)+\s\up8(→(→)=a-b+c.
[答案] a-b+c
8.如图2133所示,已知ABCDEF是一个正六边形,O是它的中心,其中,\s\up8(→(→)=b,\s\up8(→(→)=c,则\s\up8(→(→)=________.
图2133
[解析] \s\up8(→(→)=\s\up8(→(→)=\s\up8(→(→)-\s\up8(→(→)=b-c.
[答案] b-c
三、解答题
9.如图2134,解答下列各题:
(1)用a,d,e表示\s\up8(→(→);
(2)用b,c表示\s\up8(→(→);
(3)用a,b,e表示\s\up8(→(→);
(4)用d,c表示\s\up8(→(→).
图2134
[解] 因为\s\up8(→(→)=a,\s\up8(→(→)=b,\s\up8(→(→)=c,\s\up8(→(→)=d,\s\up8(→(→)=e,
所以(1)\s\up8(→(→)=\s\up8(→(→)+\s\up8(→(→)+\s\up8(→(→)=d+e+a;
(2)\s\up8(→(→)=\s\up8(→(→)-\s\up8(→(→)=-\s\up8(→(→)-\s\up8(→(→)=-b-c;
(3)\s\up8(→(→)=\s\up8(→(→)+\s\up8(→(→)+\s\up8(→(→)=a+b+e;