在数学解题中,当由条件到结论的常规定向的思考方法不能解决问题时,我们只要改变思维方向,换一个角度思考,常能找到一条绕过障碍的新的途径,构造法就是这样的一种手段,从下面两道极值题解答中,可以看到妙用构造法的巧妙之处。例1 如果实数a,b,c,d满足 a~2十b~2+2a-4b+4=0, c~2+d~2-4c+4d+4=0。求u=(a-c)~2+(b-d)~2的最值。分析:该题若用常规的代数方法会到处碰壁。换一个角度看问题,用解析法考虑:注意到(a,b),(c,d)分别是圆 (a+1)~2+(b-2)~2=1, (c-2)~2+(d-2)~2=4上的任意一点,则已知条件恰是这两圆之方程,而u=(a-c)~2+(b-d)~2恰是这两点间距离的平方,构造一个几何“模型”,在这个模型 In the mathematical problem solving, when the conventionally oriented thinking method from the condition to the conclusion cannot solve the problem, we only need to change the direction of thinking and think in a different way. We can often find a new way to bypass the obstacles. The construction method is like this. One means, from the following two extremum questions, can be seen in the clever use of the construction method. Example 1 If the real numbers a, b, c, d satisfy a~2, then b~2+2a-4b+4=0, c~2+d~2-4c+4d+4=0. Find the maximum value of u=(a-c)~2+(b-d)~2. Analysis: If you use a conventional algebraic method, the problem will hit you everywhere. Look at the problem from a different perspective and consider it analytically. Note that (a,b) and (c,d) are circles (a+1)~2+(b-2)~2=1, (c-2) respectively. For any point on ~2+(d-2)~2=4, the known condition is exactly the equation of the two circles, and u=(ac)~2+(bd)~2 is the distance between these two points. Squared, construct a geometric “model” in this model