f(x) = a - 1/x - lnxf ′(x) = 1/x² - 1/x = (1-x)/x²单调增区间：（0，1）单调减区间：（1，+∞）极大值 f(1) = a - 1/1 - ln1 = a - 1 f(x)=0恰有一个解，则f(1)=0a-1=0a=1
