\(ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{x-1}+\sqrt{\left(x+1\right)\left(x^2+1\right)}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x-1}=a\\\sqrt{x+1}=b\\\sqrt{x^2+1}=c\end{matrix}\right.\left(a,b,c\ge0\right)\), PTTT:
\(a+bc=1+abc\\ \Leftrightarrow a+bc-1-abc=0\\ \Leftrightarrow a\left(1-bc\right)-\left(1-bc\right)=0\\ \Leftrightarrow\left(a-1\right)\left(1-bc\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=1\\bc=1\end{matrix}\right.\)
Với \(a=1\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
Với \(bc=1\Leftrightarrow\sqrt{\left(x+1\right)\left(x^2+1\right)}=1\)
\(\Leftrightarrow x^3+x^2+x=0\\ \Leftrightarrow x\left(x^2+x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(ktm\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt có nghiệm duy nhất \(x=2\)
