已知函数(x-1)f(x+1/x-1)-f(x)=x,其中x不等于1,求函数解析式令a=(x+1)/(x-1)=(x-1+2)/(x-1)=1+2/(x-1)2/(x-1)=a-1x-1=2/(a-1)x=2/(a-1)+1=(a+1)/(a-1)代入(x-1)f[(x+1)/(x-1)]+f(x)=x[2/(a-1)]f(a)+f[(a+1)/(a-1)]=(a+1)/(a-1)所以[2/(x-1)]f(x)+f[
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已知函数(x-1)f(x+1/x-1)-f(x)=x,其中x不等于1,求函数解析式令a=(x+1)/(x-1)=(x-1+2)/(x-1)=1+2/(x-1)2/(x-1)=a-1x-1=2/(a-1)x=2/(a-1)+1=(a+1)/(a-1)代入(x-1)f[(x+1)/(x-1)]+f(x)=x[2/(a-1)]f(a)+f[(a+1)/(a-1)]=(a+1)/(a-1)所以[2/(x-1)]f(x)+f[
已知函数(x-1)f(x+1/x-1)-f(x)=x,其中x不等于1,求函数解析式
令a=(x+1)/(x-1)=(x-1+2)/(x-1)=1+2/(x-1)
2/(x-1)=a-1
x-1=2/(a-1)
x=2/(a-1)+1=(a+1)/(a-1)
代入(x-1)f[(x+1)/(x-1)]+f(x)=x
[2/(a-1)]f(a)+f[(a+1)/(a-1)]=(a+1)/(a-1)
所以[2/(x-1)]f(x)+f[(x+1)/(x-1)]=(x+1)/(x-1) (2)
(x-1)f[(x+1)/(x-1)]+f(x)=x (1)
(2)*(x-1)-(1)
2f(x)+(x-1)f[(x+1)/(x-1)]-(x-1)f[(x+1)/(x-1)]-f(x)=(x+1)-x
f(x)=1
我看到这种解法,但是为什么[2/(a-1)]f(a)+f[(a+1)/(a-1)]=(a+1)/(a-1)之后可以得到[2/(x-1)]f(x)+f[(x+1)/(x-1)]=(x+1)/(x-1)?
已知函数(x-1)f(x+1/x-1)-f(x)=x,其中x不等于1,求函数解析式令a=(x+1)/(x-1)=(x-1+2)/(x-1)=1+2/(x-1)2/(x-1)=a-1x-1=2/(a-1)x=2/(a-1)+1=(a+1)/(a-1)代入(x-1)f[(x+1)/(x-1)]+f(x)=x[2/(a-1)]f(a)+f[(a+1)/(a-1)]=(a+1)/(a-1)所以[2/(x-1)]f(x)+f[
[2/(a-1)]f(a)+f[(a+1)/(a-1)]=(a+1)/(a-1)---------------》》[2/(x-1)]f(x)+f[(x+1)/(x-1)]=(x+1)/(x-1)
是把前面那个式子中的a换成了x,但实际上这里的x和前面题中的x 不同.
注意!函数的表达式与字母没有关系.习惯上把自变量取为x.两个函数相等只要满足,定义域,值域,对应法则相等就可以了.