ĐK: \(x\ge1\)
PT <=> \(\sqrt{x-1}+\sqrt{\left(x^2+1\right)\left(x+1\right)}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
Ta có: \(x^3+x^2+x+1=\left(x-1\right)\left(x^2+2x+3\right)+4\ge4\)
Đặt \(\sqrt{x-1}=a\ge0;b=\sqrt{\left(x^2+1\right)\left(x+1\right)}\ge2\left(\text{theo c/m trên}\right)\) khi đó ta có:
\(a+b=1+ab\Leftrightarrow\left(a-1\right)\left(1-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\b=1\left(L\right)\end{matrix}\right.\)
Với a = 1 suy ra \(\sqrt{x-1}=1\Leftrightarrow x=2\left(TM\right)\)
P/s: Is that true?
