a . ĐKXĐ \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
P=\(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{2}{\sqrt{x}-1}+\frac{8\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x-\sqrt{x}-2\sqrt{x}-2+8\sqrt{x}-4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x+5\sqrt{x}-6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-6\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+6}{\sqrt{x}+1}\)
b. P =4\(\Leftrightarrow\frac{\sqrt{x}+6}{\sqrt{x}+1}=4\Leftrightarrow3\sqrt{x}=2\Leftrightarrow x=\frac{4}{9}\)
c. \(P>7\Leftrightarrow\frac{\sqrt{x}+6}{\sqrt{x}+1}-7>0\Leftrightarrow\frac{-\sqrt{x}-1}{\sqrt{x}+1}>0\)
\(\Leftrightarrow\sqrt{x}< -1\)vô nghiệm
Vậy không tồn tại x để P >7
d. \(P=\frac{\sqrt{x}+6}{\sqrt{x}+1}=1+\frac{5}{\sqrt{x}+1}\)
Ta thấy \(\sqrt{x}+1\ge1\Rightarrow\frac{5}{\sqrt{x}+1}\le5\Rightarrow P\le6\)
Vậy Max P =6.Dấu bằng xảy ra \(\Leftrightarrow\sqrt{x}=0\Rightarrow x=0\)
