29．已知函数f(x)=lnx−ax+1f(x)=lnx-ax+1f(x)=lnx−ax+1有两个零点x1x_{1}x1，x2x_{2}x2，且x1>2x2x_{1}>2x_{2}x1>2x2，（1）求aaa的取值范围；（2）证明：e⋅(x1+x2)>42e\cdot ({x}_{1}+{x}_{2})>4\sqrt{2}e⋅(x1+x2)>42．