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变量代换法通过式与式的相互转化,常能达到化难为易、化繁为简的目的。但在解题时极易发生下面错误,现分别举例分析如下。一、忽视原变量可取值范围,造成错误例1.若x+y+z=1,试证:x~2+y~2+z~2≥(1/3)。错解设x=(1/3)-t,y=(1/3)-2t,z=(1/3)+3t(t∈R) ∴ x~2+y~2+z~2=((1/3)-t)~2+((1/3)-2t)~2+((1/3)+3t)~2=(1/3)+14t~2≥(1/3) 当t=0,即x=y=z=1/3时,上式等号成立。剖析粗看,还以为是一个好方法,可细看,能发现其中代换x=(1/3)-t,y=(1/3)-2t,z=(1/3)+3t有欠妥当,因为x=1/ ,y=2/ ,z=4/ 显然适合已知条件x+
Variable substitution method through the mutual transformation of formula and formula, can often achieve the purpose of making things difficult, easy to simplify and simplify. However, the following errors are prone to occur when solving a problem. Examples are as follows. First, ignore the range of possible values of the original variable, resulting in errors. Example 1. If x+y+z=1, test: x~2+y~2+z~2≥(1/3). The wrong solution is set to x=(1/3)-t,y=(1/3)-2t,z=(1/3)+3t(t∈R) ∴ x~2+y~2+z~2= ((1/3)-t)~2+((1/3)-2t)~2+((1/3)+3t)~2=(1/3)+14t~2≥(1/3 When t=0, that is, x=y=z=1/3, the above equation holds. Analyze roughly and think that it is a good method. When you look closely, you can find out where substitutions x=(1/3)-t,y=(1/3)-2t,z=(1/3)+3t Not appropriate, because x=1/, y=2/, z=4/ is obviously suitable for the known condition x+