已知函数(x-1)f{(x+1)/(x-1)}+f(x)=x,其中x≠1,求函数解析式
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已知函数(x-1)f{(x+1)/(x-1)}+f(x)=x,其中x≠1,求函数解析式
已知函数(x-1)f{(x+1)/(x-1)}+f(x)=x,其中x≠1,求函数解析式
已知函数(x-1)f{(x+1)/(x-1)}+f(x)=x,其中x≠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) ,得
2f(x)+(x-1)f[(x+1)/(x-1)]-(x-1)f[(x+1)/(x-1)]-f(x)=(x+1)-x
f(x)=1,(x不等于1)
由原式得f(x)=x-(x-1)f{(x+1)/(x-1)},令x=(x+1)/(x-1),把所有的x换为(x+1)/(x-1),
得f{(x+1)/(x-1)}=(x+1)/(x-1)+{(x+1)/(x-1)-1}f{[(x+1)/(x-1)+1]/[(x+1)/(x-1)-1
化简得f{(x+1)/(x-1)}=(x+1+2f(x))/(x-1).....A
再由原式...
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由原式得f(x)=x-(x-1)f{(x+1)/(x-1)},令x=(x+1)/(x-1),把所有的x换为(x+1)/(x-1),
得f{(x+1)/(x-1)}=(x+1)/(x-1)+{(x+1)/(x-1)-1}f{[(x+1)/(x-1)+1]/[(x+1)/(x-1)-1
化简得f{(x+1)/(x-1)}=(x+1+2f(x))/(x-1).....A
再由原式变形f{(x+1)/(x-1)}=(x-f(x))/(x-1).....B
A=B,得(x+1+2f(x))/(x-1)=(x-f(x))/(x-1),化简求出f(x)=1/3,(X不等于1)
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