f(x)=√(1+x²-x)令g(x)=1+x²-x=(x-1/2)²+3/2.显然,g(x)对称轴为x=1/2,当x≥1/2时,g(x)单调递增f(x)也单调递增;x<1/2时,g(x)单调递减,f(x)也单调递减.所以“f(x)在x∈(-∞,+∞)内单调递减”不成立,原命题为假命题!