a) ta có:
\(\left\{{}\begin{matrix}1=1\\\sqrt{x}+1=\sqrt{x}+1\end{matrix}\right.\Rightarrow MTC:\sqrt{x}+1\)
Đặt \(Q\left(x\right)=\dfrac{x+\sqrt{x}}{\sqrt{x}-1}+1\)
\(Q\left(x\right)=\dfrac{x+\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}+1}{\sqrt{x}+1}\)
\(Q\left(x\right)=\dfrac{x+\sqrt{x}+\sqrt{x}+1}{\sqrt{x}+1}\)
\(Q\left(x\right)=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(Q\left(x\right)=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}=\sqrt{x}+1\)
\(\Rightarrow P\left(x\right)=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-1}.\left(\sqrt{x}+1\right)\)
\(P\left(x\right)=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}.\left(\sqrt{x}+1\right)\)
\(P\left(x\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=x-1\)
b) Thay P(x)=x-1, ta có:
\(2x^2+\left(x-1\right)=0\)
\(\Leftrightarrow x^2+x+x^2+1=0\)
\(\Leftrightarrow x\left(x+1\right)+\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=0+1=1\\x=0-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
Vậy 2x2+P(x)=0 ⇔ \(x\in\left\{-1;\dfrac{1}{2}\right\}\)