Rmo 1993 Today

( x = t^2 + 1 ), with ( t \in [2,3] ). So ( x \in [5, 10] ).

[ \sqrt(t-2)^2 + \sqrt(t-3)^2 = 1. ] That is ( |t-2| + |t-3| = 1 ), with ( t \ge 0 ). rmo 1993

Let ( t = \sqrtx-1 \ge 0 ). Then ( x = t^2 + 1 ). ( x = t^2 + 1 ), with ( t \in [2,3] )