求函数解析式

已知f [x/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),求f(x)

热心网友

已知f[x/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),求f(x)f[x/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),f(1-1/(x+1))-2f(1/(x+1))=(x+1+1)/(x+1-2)=[1+1/(x+1)]/(1-2/(x+1))用x代替上式中的1/(1+x)：即：f(1-x)-2f(x)=(1+x)/(1-2x).......(1)用1-x代替（1）式中的 x：即：f(x)-2(1-x)=(2-x)/(2x-1)........(2)(1)*2+(2):-3f(x)=[2+2x+x-2]/(1-2x)=3x/(1-2x)∴f(x)=x/(2x-1)

热心网友

f [x/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),f [1-1/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),另1/(x+1)]=u,则f [1-u]-2f（u）=（u+1）/（1-2*u）, 取值u=k,f [1-k]-2f（k）=（k+1）/（1-2*k）,(1)取值u=1-k,f（k）-2*f [1-k]=(2-k)/(2*k-1),(2)由上面两式可得：f(k)=k/(2*k-1)将上式的k值换为x,可得：f(x)=x/(2*x-1),故其解析式为：f(x)=x/(2*x-1),

热心网友

f [x/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),f [1-1/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),另1/(x+1)]=u,则f [1-u]-2f（u）=（u+1）/（1-2*u）, 取值u=k,f [1-k]-2f（k）=（k+1）/（1-2*k）,(1)取值u=1-k,f（k）-2*f [1-k]=(2-k)/(2*k-1),(2)由上面两式可得：f(k)=k/(2*k-1)将上式的k值换为x,可得：f(x)=x/(2*x-1),故其解析式为：f(x)=x/(2*x-1),

热心网友

f [x/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),f [1-1/(x+1)]-2f(1/(x+1)]=(x+2)/x-1),另1/(x+1)]=u,则f [1-u]-2f（u）=（u+1）/（1-2*u）, 取值u=k,f [1-k]-2f（k）=（k+1）/（1-2*k）,(1)取值u=1-k,f（k）-2*f [1-k]=(2-k)/(2*k-1),(2)由上面两式可得：f(k)=k/(2*k-1)将上式的k值换为x,可得：f(x)=x/(2*x-1),故其解析式为：f(x)=x/(2*x-1),