x³-3x²+6x-1=0
设 f(x) = x³-3x²+6x-1；
求导得：f '(x) = 3x² - 6x+6,
3x² - 6x+6=0
x²-2x+2=0
(x-1)²=0.
f '(x)=0 => x=1
因此(0,1)内,f '(x) > 0,所以：f(x) 在(0,1)上单调增加,
因为：f(0)=-1；f(1)=3；
故 f(x) 在(0,1)上至多只有一个零值点.
即证方程 x³-3x²+6x-1=0在区间(0,1)内只有一个实根.
Prove that there is ONLY one real root on the interval of (0
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