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法一:y=lnx是增函數,y=-2x+3是减函數,兩函數的圖像只有一個交點. f(1)=-10. 所以,函數f(x)在(1,3)上有1個零點. 法二:2x+ln x-3(x>0),f'(x)=2+1/x>0,f(x)增函數. f(1)=-10. 所以,函數f(x)在(1,3)上有1個零點.
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