\(ĐK:x\ge1\\ PT\Leftrightarrow\sqrt{2x-1}-3+\sqrt{x-1}-2=0\\ \Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}+3}+\dfrac{x-5}{\sqrt{x-1}+2}=0\\ \Leftrightarrow\left(x-5\right)\left(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\left(tm\right)\\\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}=0\end{matrix}\right.\)
Dễ thấy với \(x\ge1\) thì \(\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+2}>0\)
Vậy pt có nghiệm duy nhất \(x=5\)
Đk: \(\left\{{}\begin{matrix}2x-1\ge0\\x-1\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ge1\end{matrix}\right.\) \(\Rightarrow x\ge1\)
\(\sqrt{2x-1}+\sqrt{x-1}=5\)
\(\Rightarrow\left(\sqrt{2x-1}+\sqrt{x-1}\right)^2=25\) \(\Rightarrow2x-1+x-1+2\sqrt{2x^2-3x+1}=25\)
\(\Rightarrow2\sqrt{2x^2-3x+1}=27-3x\)
Bình phương hai vế ta được:
\(\Rightarrow4\left(2x^2-3x+1\right)=\left(27-3x\right)^2\) \(\left(đkx\le9\right)\)
\(\Rightarrow x=5\) (tmđk)
