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不妨设x>y>0,不等式化为x.ey-y.ex<x-y ,x(ey-1)<y(ex-1)即证(ey-1)/y<(ex-1)/x(x>y>0) 设f(x)=(ex-1)/x f'(x)=(ex.x-ex+1)/x2 再另g(x)=ex.x-ex+1 g'(x)=ex.x+ex-ex=ex.x>0(x>0) f'(x)min=f'(0)=0则f(x)为增函数,因为x>y则f(x)>f(
高等数学\不等式
不妨设x>y>0,不等式化为x.ey-y.ex<x-y ,x(ey-1)<y(ex-1)即证(ey-1)/y<(ex-1)/x(x>y>0) 设f(x)=(ex-1)/x f'(x)=(ex.x-ex+1)/x2 再另g(x)=ex.x-ex+1 g'(x)=ex.x+ex-ex=ex.x>0(x>0) f'(x)min=f'(0)=0则f(x)为增函数,因为x>y则f(x)>f(
高等数学\不等式
不妨设x>y>0,不等式化为x.ey-y.ex<x-y ,x(ey-1)<y(ex-1)即证(ey-1)/y<(ex-1)/x(x>y>0) 设f(x)=(ex-1)/x f'(x)=(ex.x-ex+1)/x2 再另g(x)=ex.x-ex+1 g'(x)=ex.x+ex-ex=ex.x>0(x>0) f'(x)min=f'(0)=0则f(x)为增函数,因为x>y则f(x)>f(
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