a, ĐKXĐ: x\(\ne\pm1\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}-\dfrac{8\sqrt{x}-4}{1-x}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x-1}-\dfrac{2\left(\sqrt{x}+1\right)}{x-1}+\dfrac{8\sqrt{x}-4}{x-1}\)
\(=\dfrac{x-\sqrt{x}-2\sqrt{x}-2+8\sqrt{x}-4}{x-1}\)
\(=\dfrac{x+5\sqrt{x}-6}{x-1}\)
b, \(P=\dfrac{x+5\sqrt{x}-6}{x-1}=4\)
\(\Leftrightarrow x+5\sqrt{x}-6=4\left(x-1\right)\)
\(\Leftrightarrow x+5\sqrt{x}-6=4x-4\)
\(\Leftrightarrow-3x+5\sqrt{x}-2=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(2-3\sqrt{x}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=0\\2-3\sqrt{x}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(KTMĐK\right)\\x=\dfrac{4}{9}\left(TMĐK\right)\end{matrix}\right.\)
Vậy để P=4 thì x = 4/9
c, ĐK: x-1>0 <=> x>1
\(P>7\Rightarrow\dfrac{x+5\sqrt{x}-6}{x-1}>7\)
\(\Rightarrow x+5\sqrt{x}-6>7\left(x-1\right)\)
\(\Leftrightarrow-6x+5\sqrt{x}+1>0\)
\(\Leftrightarrow\left(1-\sqrt{x}\right)\left(6\sqrt{x}+1\right)>0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}1-\sqrt{x}>0\\6\sqrt{x}+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}1-\sqrt{x}< 0\\6\sqrt{x}+1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{36}< x< 1\\x>1\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{36}< x< 1\left(TMĐK\right)\)
