ĐKXĐ : \(x-1\ge0\Leftrightarrow x\ge1\)
Ta có : \(x^4-1=\left(x^2-1\right)\left(x^2+1\right)=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)=\left(x-1\right)\left(x^3+x^2+x+1\right)\)
\(\Leftrightarrow\sqrt{x^4-1}=\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(1-\sqrt{x^3+x^2+x+1}\right)-\left(1-\sqrt{x^3+x^2+x+1}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(1-\sqrt{x^3+x^2+x+1}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}-1=0\\1-\sqrt{x^3+x^2+x+1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow x=2\left(TM\right)\)
KL.............
