1. Substitution method y = 2x+1- (x+3 under the root number) Solution: x+3=t under the root number means x = t 2-3 and t >；; =0y=2x+1- (x+3 under the radical sign) = 2 (t2-3)+1-t = 2t2-t-5 = 2 (t-1/2) 2-5-65433. =0 The evaluation domain of quadratic function is obvious y & gt=- 1 1/2, so the value range is [- 1 1/2, positive infinity) 2. Matching method y = x 4+2x 2-1solution: y = (x 2+65438+) infinity) 3. Separation method F (X) = X+ 1×- 1 solution: f (x) = 4 (x+1)-5/x+1= 4-(5/x+65438+) 0.