Rút gọn :
P = \(\left(\frac{3\sqrt{x}-1}{x-1}-\frac{1}{\sqrt{x}-1}\right):\frac{1}{x+\sqrt{x}}\)
= \(\left(\frac{3\sqrt{x}-1}{x-1}-\frac{1\left(\sqrt{x}+1\right)}{x-1}\right).\sqrt{x}\left(\sqrt{x}+1\right)\)
= \(\frac{3\sqrt{x}-1-\sqrt{x}-1}{x-1}.\sqrt{x}\left(\sqrt{x}+1\right)\)
= \(\frac{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{x-1}\)
= \(2\sqrt{x}\)
a)Rgọn P=\(\left(\frac{3\sqrt{x}-1}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{x+\sqrt{x}}\)
= \(\frac{3\sqrt{x}-1-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\sqrt{x}\left(\sqrt{x}+1\right)\)
= \(\frac{2\left(\sqrt{x}-1\right)\sqrt{x}}{\sqrt{x}-1}\)
= 2\(\sqrt{x}\)
b) Ta có : \(2P-x=3\)
=> \(4\sqrt{x}-x=3\)
=> \(-\left(x-4\sqrt{x}+3\right)=0\)
=> \(-\left(x-\sqrt{x}\right)\left(3\sqrt{x}-3\right)=0\)
=> \(-\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)=0\)
=> \(\left(\sqrt{x}-1\right)\left(-\sqrt{x}-3\right)=0\)
=> \(\left[{}\begin{matrix}\sqrt{x}-1=0\\-\sqrt{x}-3=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=1\left(loại\right)\\x=9\end{matrix}\right.\)
Vậy để 2P -x=3 thì x =9
