a, *) ĐXKĐ: \(x>0\).
\(P=\frac{\sqrt{x}\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\frac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\\ =\frac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\left(2\sqrt{x}+1\right)+1\\ =\sqrt{x}\left(\sqrt{x}+1\right)-2\sqrt{x}-1+1\\ =x+\sqrt{x}-2\sqrt{x}\\ =x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\)
*) Để P=2 thì:
\(x-\sqrt{x}=2\\ \Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow x-2\sqrt{x}+\sqrt{x}-2=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}-2\right)+\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=0\\ \left[{}\begin{matrix}\sqrt{x}+1=0\\\sqrt{x}-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(vn\right)\\\sqrt{x}=2\end{matrix}\right.\\ \Leftrightarrow x=4\left(t/m\right)\)
b, Với \(x>1\) thì \(\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}>1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}>0\\\sqrt{x}-1>0\end{matrix}\right.\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=P>0\)
Suy ra \(\left|P\right|=P\), hay \(P-\left|P\right|=0\).
Chúc bạn học tốt nha
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