①h(x)=f(x)-g(x)=ln(e^x+1)-kx,偶函数h(-x)=ln[e^(-x)+1]+kx=ln(e^x+1)-kx=h(x)=> ln[(1+e^x)/e^x]+kx=ln(e^x+1)-kx=> ln(e^x+1)-x+kx=ln(e^x+1)-kx=> -x+kx=-kx=> k=1/2②√(x-1)0=> [x+2(t-1)]^2+4(t^2+1)-4(t-1)...