\(a,P=\dfrac{\sqrt{x}+1+x}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\\ P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\\ b,x=\dfrac{8}{\sqrt{5}-1}-\dfrac{8}{\sqrt{5}+1}=\dfrac{8\left(\sqrt{5}+1-\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\\ x=\dfrac{16}{4}=4\Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow P=\dfrac{4+2+1}{2}=\dfrac{7}{2}\\ c,P=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\\ \ge2\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}+1=2\cdot1+1=3\\ \Leftrightarrow P\ge3\\ d,P=2\sqrt{x}-2\Leftrightarrow x+\sqrt{x}+1=2x-2\sqrt{x}\\ \Leftrightarrow x-3\sqrt{x}-1=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{3+\sqrt{5}}{2}\\\sqrt{x}=\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7+3\sqrt{5}}{2}\\x=\dfrac{7-3\sqrt{5}}{2}\end{matrix}\right.\)
\(e,\) Vì \(P\ge3>0\) nên \(\left|P\right|=P\)
Do đó \(M=P-P=0\)
\(f,\) Từ câu c ta được \(P_{min}=3\Leftrightarrow\sqrt{x}=\dfrac{1}{\sqrt{x}}\Leftrightarrow x=1\left(tm\right)\)
