求下列各题中的函数f(x)的解析式

(1)t=√x +2 t>=2 √x=t-2 f(t)=(t-2)²+4(t-2)
(2)f(x+1)=(x+1)²-5(x+1)+6 =>f(x)=x²-5x+6
(3)x=1/t 2f(1/t)+f(t)=2/t
<=>2f(1/x)+f(x)=2/x-------------------(1)
2f(x)+f(1/x)=2x--------------------(2)
联立（1）（2）解得f(x)=1/3(4x-2/x)=2/3(2x-1/x)
(4)f(x)=ax+b
f[f(x)]=a²x+ab+b
a²=4 b(a+1)=-1
=>a=2 b=-1/3
或a=-2 b=1

(1)(√x +2 )²=x+4√x+4
x+4√x=(√x +2 )²-4
f(√x +2 )=x+4√x=(√x +2 )²-4
f(x)=x²-4
(2)f(x+1)
=x^2-3x+2
=x^2+2x-5x+1+1
=(x+1)^2-5x+1
=(x+1)^2-5x-5+6
=(x+1)^2-5(x+1)+6
f(x)=x^2-5x+6
(3)2f(x)+f(1/x)=2x
4f(x)+2f(1/x)=4x
2f(1/x)+f(x)=2/x
3f(x)=4x-2/x
f(x)=(4x-2/x)/3
(4)f(x)=kx+b
f((f(x))=k(kx+b)+b=k^2x+bk+b=4x-1
k^2=4
bk+b=-1
k=2 b=-1/3
k=-2 b=1
f(x)=2x-1/3
f(x)=-2x+1

1、f(x)=x^2-4
2、f(x)=x^2-5x
3、f(X)=(4/3)x-2/(3x)
4、f(x)=-2x+1或=2x-1/3
1、2、3换元法，4设f(x)=ax+b