√x和√(x+a)在定义域内均未增函数,所以f(x)在定义域左端取最小值.(i) a ≥ 0f(x)的定义域:[0,+∞)f(x)最小值 = f(0) = √a = √3/2,a = 3/4(ii) a < 0f(x)的定义域:[-a,+∞)f(x)最小值 = f(-a) = √(-a) + √(-a + ...