ĐK: \(-1\le x\le1\)
Pt đã cho \(\Leftrightarrow\)\(\sqrt{x}\left(\sqrt{x+1}+\sqrt{1-x}\right)=x+1\Rightarrow x\left(2+2\sqrt{1-x^2}\right)=x^2+2x+1\Rightarrow2+2\sqrt{1-x^2}=x+2+\dfrac{1}{x}\Leftrightarrow2\sqrt{1-x^2}=x+\dfrac{1}{x}\Rightarrow4-4x^2=x^2+\dfrac{1}{x^2}+2\Leftrightarrow5x^2+\dfrac{1}{x^2}-2=0\Rightarrow5x^4-2x^2+1=0\)(x>0)
Vì pt \(5x^4-2x^2+1\) vô nghiệm nên pt ban đầu cũng vô nghiệm