1、y=√(x+1)
∵x+1≥0
∴y=√(x+1)≥0
2、y=(1-x²)/(1+x²)
∵x∈R,y(1+x²)=(1-x²)即(y+1)x²+(y-1)=0
∴Δ=0-4(y+1)(y-1)≥0
解得：-1≤y≤1
求值域：1、y=根号x+1 2、y=(1-x²)/(1+x²)
1、y=√(x+1)
∵x+1≥0
∴y=√(x+1)≥0
2、y=(1-x²)/(1+x²)
∵x∈R,y(1+x²)=(1-x²)即(y+1)x²+(y-1)=0
∴Δ=0-4(y+1)(y-1)≥0
解得：-1≤y≤1