B10:
a.đk : \(x\ge0\) ; \(x\ne1\)
\(P=\left[\dfrac{\sqrt{x}\left(x+1\right)}{2\left(x-1\right)}+\dfrac{3-\sqrt{x}}{2\left(x-1\right)}\right]:\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+2}{x\sqrt{x}-1}\right]\)
\(P=\left(\dfrac{x+\sqrt{x}+3-\sqrt{x}}{2\left(x-1\right)}\right):\left(\dfrac{x-1+\sqrt{x}+2}{x\sqrt{x}-1}\right)\)
\(P=\dfrac{x+3}{2x-1}:\dfrac{1}{\sqrt{x}-1}=\dfrac{x+3}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\dfrac{\sqrt{x}-1}{1}\)
\(P=\dfrac{\left(x+3\right)\left(\sqrt{x}-1\right)}{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x+3}{2\left(\sqrt{x}+1\right)}=\dfrac{x+3}{2\sqrt{x}+2}\)
Có: \(\left|x+2\sqrt{x}=3\right|\Leftrightarrow x+2\sqrt{x}=3.do.x\ge0\)
=> \(x+\sqrt{x}-3=0\)
<=> \(\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)=0\)
<=> \(\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\left(loại\right)\\\left(loại\right)\end{matrix}\right.\)
=> Không tồn tại x
=> P không xác định được giá trị khi \(\left|x+2\sqrt{x}=3\right|\)
#TueLam
