将(3)代入(①)式中求c值 k1+k2 (a-b(a-c)=k, (a-b) 从而 a k+bk, k1+k2 k,+ 故取A k2 ak+bk2即可满足要求 (a-b) k1+k2将(3)代入 (1)式中求c 值: ( )( ) , ( ) 2 1 1 2 a b a c k a b k k − − = − + 从而 . 1 2 1 2 k k a k b k c + + = , , . ( ) 1 2 1 2 2 故取 1 2 即可满足要求 k k a k b k c a b k k A + + = − + =