1.2x+1≥0,得x≥-1/2,定義域[-1/2,+∞)
y=x+2√(2x+1)
即x^2-2(y+4)x+y^2-4=0
設f(x)=x^2-2(y+4)x+y^2-4=[x-(y+4)]^2-8y-20
當y+4≤-1/2,即y≤-9/2時,要使得f(x)在[-1/2,+∞)有與x軸有交點
則f(-1/2)≤0,即(1/4)+(y+4)+y^2-4≤0,得y=-1/2,與y≤-9/2矛盾
當y+4>-1/2,即y>-9/2時要使得f(x)在[-1/2,+∞)有與x軸有交點
則f(y+4)≤0,即-8y-20≤0,得y≥-5/2
所以值域[-5/2,+∞)
2.x>0,-x