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證明:設f(x)=x^3-3x-1,則f'(x)=3x^2-3 ∵x>1,∴x^2>1,∴3x^2-3>0 即f'(x)>0,∴函數f(x)在(1,2)上單調遞增 而f(1)=-10 ∴f(x)至少與x軸有一個交點 即方程x^3-3x=1在(1,2)內至少有一個實根 望採納!有問題請追問!
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