`\sqrt(x+1)+x^2=\sqrt(1-y)+y^2`
`⇔\sqrt(x+1)+x^2-\sqrt(1-y)-y^2=0`
`⇔(x+1-1+y)/(\sqrt(x+1)+\sqrt(1-y))+x^2-y^2=0`
`⇔(x+y)(1/(\sqrt(x+1)+\sqrt(1-y))+x-y)=0`
`th1 `
`x+y=0`
`⇔x=-y`
`''=''` khi
`⇒x=0`
`⇒x=y=0`
`th2`
`1/(\sqrt(x+1)+\sqrt(1-y))+x-y=0`
vì `1/(\sqrt(x+1)+\sqrt(1-y))>0`
`⇒x-y<0`
`⇔x<y`
mà `y≤1`
`x≥1`
`⇒x≥y`
nhưng
`x<y`
`⇒`loại
`⇒x=y=0`là nghiệm duy nhất
