复合函数：1.已知f(x)=根号x，g(x)=sin(x),求f[g(x)],g[f(x)] 2.已知f(x)=(x-1)/(x+1),求f[f(x)]

f(x)=√x，g(x)=sin(x)

f(g(x))=f(sin(x))=√sinx
g(f(x))=g(√x))=sin(√x)

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

f[g(x)]=根号sin(x) , g[f(x)]=sin(根号x),

f[f(x)]={(x-1)/(x+1)-1]/[(x-1)/(x+1)+1]
=-1/x

1=f<g(x)>=根号下sinx
g<f(x)>=sin根号下x
f<f(x)>=<(x-1)(x+1)-1><(x-1)(x+1)+1>然后化简。