设函数g(x)=(x+1)ln(x+1)-x,f(x)=a(x+1)^2ln(x+1)+bx,曲线y=f(x)在原点(0,0)处的切线方程为y=0,且经过点(e-1,e^2-e+1)(1)求y=f(x)表达式,并证明:当x》0时,g(x)》0(2)若当x》0时,f(x
设函数g(x)=(x+1)ln(x+1)-x,f(x)=a(x+1)^2ln(x+1)+bx,曲线
y=f(x)在原点(0,0)处的切线方程为y=0,且经过点(e-1,e^2-e+1)
(1)求y=f(x)表达式,并证明:当x》0时,g(x)》0
(2)若当x》0时,f(x)》mx^2恒成立,求实数m的取值范围