ĐKXĐ: \(x\ge1\)
\(\left(x-1\right)^2+2\sqrt{x-1}\left(x-1\right)+\left(x-1\right)-4=0\)
\(\Leftrightarrow\left(x-1+\sqrt{x-1}\right)^2-2^2=0\)
\(\Leftrightarrow\left(x-3+\sqrt{x-1}\right)\left(x+1+\sqrt{x-1}\right)=0\) (1)
Do \(x+1+\sqrt{x-1}>0\) \(\forall x\ge1\)
\(\left(1\right)\Leftrightarrow x-3+\sqrt{x-1}=0\Leftrightarrow\left\{{}\begin{matrix}3-x\ge0\\x-1=\left(3-x\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le3\\x^2-7x+10=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=5\left(l\right)\end{matrix}\right.\)
Vậy pt có nghiệm duy nhất \(x=2\)
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