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一些求值问题设字母替换来解,方法别具一格。今举例给予说明。例1 求(2+3~(1/2))~(1/2)+(2-3~(1/2))~(1/2)的值。解:设x=(2+3~(1/2))~(1/2)+(2-3~(1/2))~(1/2)则 x~2=((2+3~(1/2))~(1/2)+(2-3~(1/2))~(1/2))~2=2+3~(1/2)+2(((2+3~(1/2))((2-3~(1/2)))~(1/2)+2-3~(1/2) =4+2(2~2-(3~(1/2))~2)=6 ∴x=±6~(1/2) (-(6~(1/2))不合题意舍去) 因此,原式=6~(1/2)。例2 求 (4-3((4-3((4-3)~(1/3))~(1/3)))~(1/3)…的值。解:设x=(4-3((4-3((4-3)~(1/3))~(1/3)))~(1/3) 则 x~3=4-3((4-3((4-3~(1/3))))~(1/3)…即 x~3=4-3x。∴x=1注:例2应该先证其存在性之后才能设,这
Some of the evaluation problems are based on the replacement of letters to solve the problem. This example gives instructions. Example 1 Find the value of (2+3~(1/2))~(1/2)+(2-3~(1/2))~(1/2). Solution: Let x=(2+3~(1/2))~(1/2)+(2-3~(1/2))~(1/2) then x~2=((2+3 ~(1/2))~(1/2)+(2-3~(1/2))~(1/2))~2=2+3~(1/2)+2(((2 +3~(1/2))((2-3~(1/2)))~(1/2)+2-3~(1/2)=4+2(2~2-(3~ (1/2))~2)=6 ∴x=±6~(1/2) (-(6~(1/2)) is not rounded off) Therefore, the original formula = 6~(1/2) Example 2 Find the value of (4-3((4-3((4-3)~(1/3))~(1/3)))~(1/3)... Solution: Let x= (4-3((4-3((4-3)~(1/3))~(1/3)))~(1/3) then x~3=4-3((4-3( (4-3~(1/3))))~(1/3)... That is, x~3=4-3x.∴ x=1 Note: Example 2 should be established after verifying its existence.