换元法是中学数学的一种基本解题方法,使用这种方法常使得要解决的问题由繁变简,化难为易。下面以求函数最大(小)值为例,指出使用换元法时应注意的几个问题。一、在换元时要注意变量的允许值范围例1若x+y=1,求S=(x-3)~2+y~2的最大值和最小值。见到条件x+y=1,学生常会设x=sin~2α,y=cos~2α,从而化得S=2(sin~2α-2)~2+2,故有S_(max)=10,S_(min)=4。这个结论显然是错误的。错误的原因在于换元时未注意到变量的允许值范围应保持不变,由题目的条件,变量x、y可以取任意实数值(只要满足x+y=1即可),但换元后0≤x=sin~2α≤1,0≤y=cos~2α≤1,可见,允许值范围发生了变化。使用换元法,例1可以这样解: 设x=2+t,y=-1-t(t为任意实数),则 Redemption method is a basic problem-solving method in middle school mathematics. Using this method often makes the problem to be solved simpler and harder. The following function to find the largest (small) value, for example, pointed out that the use of replacement method should pay attention to several issues. First, pay attention to the allowable value range of variables when changing yuan Example 1 If x + y = 1, find the maximum and minimum S = (x-3) ~ 2 + y ~ 2. See the condition x + y = 1, students often set x = sin ~ 2α, y = cos ~ 2α, so that the S = 2 (sin ~ 2α-2) ~ 2 +2, so there S max = , S_ (min) = 4. This conclusion is obviously wrong. The reason for the error is that the variable does not change the allowable value range should be unchanged, the subject of the conditions, variables x, y can take any real value (as long as x + y = 1 can be), but after the swap 0≤x = sin ~ 2α≤1, 0≤y = cos ~ 2α≤1, it can be seen that the allowable value range has changed. Using the method of commutation, Example 1 can be solved in this way: Let x = 2 + t and y = -1-t (t is an arbitrary real number), then