1．用二次根式被开方数的非负性进行夹逼例1 已知x是实数,则的值是( ) (A)1-(1/π)． (B)1+(1/π)． (C)(1/π)-1．(D)无法确定． (第十四届“希望杯”) 解根据二次根式被开方数的非负性知, x-π≥0,且π-x≥0, 即 x≥π,且x≤π, 所以 x=π．从而原式=0+0+(π-1)/π=1-(1/π), 故选(A)． 2．用两非负数和为零进行夹逼 1. Using quadratic roots to be clamped by the non-negative nature of the number of square roots. Example 1 Given that x is a real number, the value is () (A)1-(1/π). (B) 1+(1/π). (C)(1/π)-1. (D) Unable to determine. (The 14th “Hope Cup”) solution based on the non-negative knowledge of the square root of the squared formula, x-π ≥ 0, and π-x ≥ 0, that is x ≥ π, and x ≤ π, so x =π. Thus the original formula=0+0+(π-1)/π=1-(1/π), so choose (A). 2. Clamp between two non-negative numbers and zero