a.
ĐKXĐ: \(x\ge2\)
\(\left(x+\sqrt{x}+1\right)\sqrt{x-2}=\left(x+1\right)^2-x\)
\(\Leftrightarrow\left(x+\sqrt{x}+1\right)\sqrt{x-2}=\left(x+\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x-2}=x-\sqrt{x}+1\)
\(\Leftrightarrow\sqrt{x-2}+\sqrt{x}=x+1\)
\(\Leftrightarrow2x-2+2\sqrt{x^2-2x}=x^2+2x+1\)
\(\Leftrightarrow x^2-2\sqrt{x^2-2x}+3=0\)
\(\Leftrightarrow\left(\sqrt{x^2-2x}-1\right)^2+2x+2=0\) (vô nghiệm do \(2x+2>0\))
Vậy pt đã cho vô nghiệm
b. ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow2x^2-3x+1+2\left(x-1\right)\sqrt{2x^2-3x+1}+x^2-2x-3=0\)
Đặt \(\sqrt{2x^2-3x+1}=t\ge0\)
\(\Rightarrow t^2+2\left(x-1\right)t+x^2-2x-3=0\)
\(\Delta'=\left(x-1\right)^2-\left(x^2-2x-3\right)=4\)
\(\Rightarrow\left[{}\begin{matrix}t=1-x-2=-x-1\\t=1-x+2=3-x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x^2-3x+1}=-x-1\left(x\le-1\right)\\\sqrt{2x^2-3x+1}=3-x\left(x\le3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x=0\left(vn\right)\\x^2+3x-8=0\left(x\le3\right)\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{-3\pm\sqrt{41}}{2}\)