đk : \(x\ge1\)
\(\Leftrightarrow x\sqrt{x}=\sqrt{x^2-1}+\sqrt{x-1}\)
\(\Leftrightarrow x\sqrt{x}=\sqrt{x-1}\left(\sqrt{x+1}+1\right)\)
\(\Leftrightarrow x\sqrt{x}=\sqrt{x-1}.\frac{\left(x+1\right)-1}{\sqrt{x+1}-1}\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}-1\right)=\sqrt{x-1}\)( ví \(x\ge1>0\))
\(\Leftrightarrow x\left(x+2-2\sqrt{x+1}\right)=x-1\)( vì \(x\ge1\)nên \(\sqrt{x+1}-1>0\))
\(\Leftrightarrow x^2+x+1-2x.\sqrt{x+1}=0\)
\(\Leftrightarrow x^2-2x\sqrt{x+1}+\left(x+1\right)=0\)( ta có thể lập pt 2 vế )
\(\Leftrightarrow x-\sqrt{x+1}=0\Leftrightarrow x=\sqrt{x+1}\Leftrightarrow x^2=x+1\)
\(\Leftrightarrow x^2-x-1=0\Leftrightarrow x=\frac{1+\sqrt{5}}{2}\)hoặc \(x=\frac{1-\sqrt{5}}{2}\)
\(\Leftrightarrow x=\frac{1+\sqrt{5}}{2}\)( vì đk \(x\ge1\))
Vậy nghiệm của pt là \(x=\frac{1+\sqrt{5}}{2}\)
