求函数y=2-√(-x²-3x+4)的单调区间定义域：由-x²-3x+4≧0,得x²+3x-4=(x+4)(x-1)≦0,故定义域为-4≦x≦1.设y=2-√u,u=-x²-3x+4=-(x²+3x)+4=-[(x+3/2)²-9/4]+4=-(x+3/2)²+25/4≦25...