f(x)=(X^2+8)/(X-1)
=(X^2-2X+1+2X-2+9)/(X-1)
=(X^2-2X+1)/(X-1)+(2X-2)/(X-1)+9/(X-1)
=(X-1)+2+9/(X-1)
=[√(X-1)-3/√(X-1)]^2+6+2
=[√(X-1)-3/√(X-1)]^2+8
>=8
所以函数f(x)=(X^2+8)/(X-1)的最小值为8