\(M=\left(\dfrac{3\sqrt{x}-1}{x-1}-\dfrac{1}{x+\sqrt{x}}\right)\div\dfrac{1}{x+\sqrt{x}}\)
\(=\left[\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\times\sqrt{x}\left(\sqrt{x}+1\right)\)
\(=\dfrac{\sqrt{x}\left(3\sqrt{x}-1\right)}{\sqrt{x}-1}-1\)
\(=\dfrac{3x-\sqrt{x}-\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{3x-2\sqrt{x}+1}{\sqrt{x}-1}\)
\(2P-x=3\Leftrightarrow2\left(\dfrac{3x-2\sqrt{x}+1}{\sqrt{x}-1}\right)-x=3\)
\(\Leftrightarrow\dfrac{6x-4\sqrt{x}+2-x\sqrt{x}+x-3\sqrt{x}+3}{\sqrt{x}-1}=0\)
\(\Leftrightarrow7x-7\sqrt{x}-x\sqrt{x}+5=0\)
Đến đây sao số xấu quá (T^T)
a) với x>0, x#1 ta có
\(P=\left(\dfrac{3\sqrt{x}-1}{x-1}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{1}{x+\sqrt{x}}\)
\(=\left(x+\sqrt{x}\right)\left(\dfrac{3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\sqrt{x}\left(\sqrt{x}+1\right).\dfrac{3\sqrt{x}-1-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\sqrt{x}\left(2\sqrt{x}-2\right)}{\sqrt{x}-1}=\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}=2\sqrt{x}\)
vậy với x>0 , x#1 thì P=\(2\sqrt{x}\)
b)vậy với x>0 , x#1 và P=\(2\sqrt{x}\) ta có
\(2P-x=3\Leftrightarrow4\sqrt{x}-x=3\\ \Leftrightarrow x-4\sqrt{x}+3=0\\ \Leftrightarrow x-\sqrt{x}-3\sqrt{x}+3=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)-3\left(\sqrt{x}-1\right)=0\\ \Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}-1=0\\\sqrt{x}-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\left(loại\right)\\x=9\left(nhận\right)\end{matrix}\right.\)
vậy với x=9 thì 2P-x=3