f(x)=e^x(e^x-2a- 1)-(a+ 1)x
f'=e^x(e^x-2a- 1)+e^2x-(a+ 1)
=e^x(2e^x-2a- 1)-(a+ 1)=2(e^x)？ -(2a+ 1)e^x-(a+ 1)
Taking e x as a variable is a quadratic equation with one variable.
δ=(2a+ 1)？ +8(a+ 1)
=4a？ +4a+ 1+8a+8
=4a？ + 12a+9
=(2a+3)？ ≥0
Root e x = [2a+ 1 (2a+3)]/4
e^x 1=[2a+ 1+(2a+3)]/4=a+ 1；
Ex2 = [2a+1-(2a+3)]/4 =-1/2 (impossible);
(1)a+ 1≤0, a≤- 1, f'(x)=0 has no solution, and f' (x) > 0; 0, which is an increasing function in all fields r.
(2)a & gt； -1, f'(x)=0 has roots, e x = a+ 1, x = ln (a+ 1),
X≥ln(a+ 1), increasing function; X & ltLn(a+ 1), subtraction function;
When x = ln (a+ 1) and e x = a+ 1, there is a minimum value.
(ii) Minimum value >; 0 will do.
A≤- 1, which is an increasing function in all fields r,
f(-∞)& gt； 0,
f(-∞)= 0-(a+ 1)(-∞)=(a+ 1)∞，a+ 1 & gt； 0, a>- 1, contradiction, no solution;
a & gt- 1，f(x)min = f(ln(a+ 1))=(a+ 1)(a+ 1-2a- 1)-(a+ 1)ln(a+ 1)
=-a(a+ 1)-(a+ 1)ln(a+ 1)
=-(a+ 1)(a+ln(a+ 1))& gt； 0
a+ln(a+ 1)& lt； 0,
h(a)=a+ln(a+ 1)
H' (a) =1+1(a+ 1) > 0, h(a) is an increasing function, and a+1
∴ The solution is a>0