Cách giải:
`x^2+\sqrt{1+x}+\sqrt{1-x}-2=0`
`ĐK:-1<=x<=1`
`pt \harr \sqrt{1+x}+\sqrt{1-x}=2-x^2`
`->(\sqrt{1+x}+\sqrt{1-x})^2=(2-x)^2`
`->2+2\sqrt{1-x^2}=(1-x^2+1)^2`
Đặt `\sqrt{1-x^2}=y(y>=0)->1-x^2=y^2`
`->2+2y=(y^2+1)^2`
`->y^4+2y^2+1-2y-2=0`
`->y^4-1+2y(y-1)=0`
`->(y-1)(y^3+y^2+3y+1)=0`
`->y=1`
`->x^2=0`
`->x=0(TM)`
Vậy pt có nghiệm x=0