Đáp án và giải thích các bước giải:
`ĐK` : `-1≤x≤1`
`\sqrt[x+1]+2(x+1)=x-1+\sqrt[1-x]+3\sqrt[1-x^2]`
`=\sqrt[1+x]+2(x+1)=x-1+\sqrt[1-x]+3\sqrt[(1-x)(1+x)]` `(1)`
Đặt `\sqrt[1+x]=u` , `\sqrt[1-x]=t` `(u,t>0)`
`(1)⇔u+2u^2=-t^2+t+3ut`
`⇔` `(u-t)^2+u(u-t)+(u-t)=0`
`⇔` `(u-t)(u-t+u+1)=0`
`⇔` `(u-t)(2u-t+1)=0`
`⇔` \(\left[ \begin{array}{l}u=t\\2u+1=t\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\4(x+1)+4\sqrt{x+1}+1=1-x\end{array} \right.\)
`⇔` \(\left[ \begin{array}{l}x=0\\4\sqrt{x+1}=-(5x+4)(*)\end{array} \right.\)
`(*)⇔4\sqrt[x+1]=-(5x+4)` `(-1≤x≤-5/4)`
`⇔` `25x^2+24x=0`
`⇔` `x(25x+24)=0`
`⇔` \(\left[ \begin{array}{l}x=0\\x=\dfrac{-24}{25}\end{array} \right.\)
Vậy `S={0,-{24}/{25}}`
