\(ĐK:x\ge\dfrac{1}{2}\\ PT\Leftrightarrow\left(5x-5\right)-\left(2\sqrt{2x^2+5x-3}-4\right)-\left(x\sqrt{2x-1}-x\right)+\left(2x\sqrt{x+3}-4x\right)=0\\ \Leftrightarrow5\left(x-1\right)-\dfrac{2\left(2x+7\right)\left(x-1\right)}{\sqrt{2x^2+5x-3}+2}-\dfrac{x\left(2x-2\right)}{\sqrt{2x-1}+1}+\dfrac{2x\left(x-1\right)}{\sqrt{x+3}+2}=0\\ \Leftrightarrow\left(x-1\right)\left(5-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+2}-\dfrac{x}{\sqrt{2x-1}+1}+\dfrac{2x}{\sqrt{x+3}+2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\5-\dfrac{2\left(2x+7\right)}{\sqrt{2x^2+5x-3}+2}-\dfrac{x}{\sqrt{2x-1}+1}+\dfrac{2x}{\sqrt{x+3}+2}=0\left(1\right)\end{matrix}\right.\)
Với \(x\ge\dfrac{1}{2}\Leftrightarrow\left(1\right)< 0\)
Do đó PT có nghiệm x=1
ĐKXĐ: \(x\ge\dfrac{1}{2}\)
\(2x-2\sqrt{\left(2x-1\right)\left(x+3\right)}-\left(1+x\sqrt{2x-1}-2x\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left(2x-1\right)-2\sqrt{\left(2x-1\right)\left(x+3\right)}-x\sqrt{2x-1}+2x\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{2x-1}\left(\sqrt{2x-1}-x\right)-2\sqrt{x+3}\left(\sqrt{2x-1}-x\right)=0\)
\(\Leftrightarrow\left(\sqrt{2x-1}-x\right)\left(\sqrt{2x-1}-2\sqrt{x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=x\\\sqrt{2x-1}=2\sqrt{x+3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x^2\\2x-1=4x+12\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\2x=-13\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-\dfrac{13}{2}\left(ktm\right)\end{matrix}\right.\)
