\(ĐKXĐ:x>0;x\ne1\)
\(a.M=\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x+1}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}=\dfrac{x+\sqrt{x}+1-x+\sqrt{x}-1+x+1}{\sqrt{x}}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(b.M=\dfrac{9}{2}\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-\dfrac{9}{2}=0\)
\(\Leftrightarrow\dfrac{2x+4\sqrt{x}+2-9\sqrt{x}}{2\sqrt{x}}=0\)
\(\Leftrightarrow2x-\sqrt{x}-4\sqrt{x}+2=0\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(TM\right)\\x=\dfrac{1}{4}\left(TM\right)\end{matrix}\right.\)
\(c.\) Xét : \(M\ge4\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-4\ge0\)
\(\Leftrightarrow\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\ge0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\ge0\left(luôn-đúng\right)\)
KL......