a) Ta có: \(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\left(x\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}+\frac{\left(x+1\right)\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x^2+x\sqrt{x}-\sqrt{x}-1-x^2+x\sqrt{x}-\sqrt{x}+1+x^2-1}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{x^2+2x\sqrt{x}-2\sqrt{x}-1}{\sqrt{x}\cdot\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{\left(x-1\right)\left(x+1\right)+2\sqrt{x}\left(x-1\right)}{\sqrt{x}\cdot\left(x-1\right)}\)
\(=\frac{\left(x-1\right)\left(x+2\sqrt{x}+1\right)}{\sqrt{x}\cdot\left(x-1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
Để \(M=\frac{9}{2}\) thì \(\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac{9}{2}\)
\(\Leftrightarrow2\left(\sqrt{x}+1\right)^2=9\sqrt{x}\)
\(\Leftrightarrow2\left(x+2\sqrt{x}+1\right)=9\sqrt{x}\)
\(\Leftrightarrow2x+4\sqrt{x}+2-9\sqrt{x}=0\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow2x-4\sqrt{x}-\sqrt{x}+2=0\)
\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2=0\\2\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2\\2\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\\sqrt{x}=\frac{1}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\frac{1}{4}\left(nhận\right)\end{matrix}\right.\)
Vậy: Khi \(M=\frac{9}{2}\) thì \(x\in\left\{4;\frac{1}{4}\right\}\)
c) Ta có: \(\left(\sqrt{x}+1\right)^2\ge4\cdot\sqrt{x}\)
\(\Leftrightarrow\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\ge\frac{4\cdot\sqrt{x}}{\sqrt{x}}=4\)
hay \(M\ge4\)
