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有这么一道函数方程:若f(x)-2f(1/x)=x,求f(x)。有这样一种解法: ∵ f(x)-2f(1/x)=x=(-3x~2/-3x)=(x~2+2-3-4x~2)/-3x=(x~2+2)/(-3x)-(2·(1+2x~2)/-3x)=(x~2+2)/(-3x)-(2·(1/x~2)+2/(-3·1/x)。∴ f(x)-(x~2+2)/(-3x)。我真不知道怎么想到要把x写成(-3x~2)/(-3x)?更不知道怎么想到要把-3x~2写成x~2+2+2-4x~2?我疑心的是编者事先求出f(x),再倒过来创造了这种解法的。我知道的是,要从f(x)-2f(1/x)=x中求出f(x)。于是要想办法消去f(1/x),因此还得
There is such a function equation: If f(x)-2f(1/x)=x, find f(x). There is such a solution: ∵ f(x)-2f(1/x)=x=(-3x~2/-3x)=(x~2+2-3-4x~2)/-3x=(x ~2+2)/(-3x)-(2(1+2x~2)/-3x)=(x~2+2)/(-3x)-(2(1/x~2)+ 2/(-3·1/1/x) ∴ f(x)-(x~2+2)/(-3x) I really don’t know how to think of x as (-3x~2)/(-3x I don’t know how to think about writing -3x~2 as x~2+2+2-4x~2? My doubt is that the editor had to find f(x) in advance, and invert it to create this kind of solution. What we know is that we need to find f(x) from f(x)-2f(1/x)=x, so we have to eliminate f(1/x), so we have to