y=√(x^2-1)
y'=x/√(x^2-1)
y'=[x√(x^2-1)]/(x^2-1)
令：y'＜0,即：[x√(x^2-1)]/(x^2-1)＜0
整理,有：x√(x^2-1)＜0
解得：x＜0
同时,因x^2-1≥0,解得：x≤-1或x≥1
因此,所给函数单调递减的区间是∈(-∞,-1]
函数y=根号下(x^2-1)的单调递减区间为
y=√(x^2-1)
y'=x/√(x^2-1)
y'=[x√(x^2-1)]/(x^2-1)
令：y'＜0,即：[x√(x^2-1)]/(x^2-1)＜0
整理,有：x√(x^2-1)＜0
解得：x＜0
同时,因x^2-1≥0,解得：x≤-1或x≥1
因此,所给函数单调递减的区间是∈(-∞,-1]