a) \(M=\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}\right)^3-1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}\right)^3+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\sqrt{x}}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}+\frac{x+1}{\sqrt{x}}\)
\(=\frac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)+\left(x+1\right)}{\sqrt{x}}\)
\(=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) Để \(M=\frac{9}{2}\) thì :
\(\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac{9}{2}\Leftrightarrow2\left(\sqrt{x}+1\right)^2=9\sqrt{x}\)
\(\Leftrightarrow2x+4\sqrt{x}+2-9\sqrt{x}=0\)
\(\Leftrightarrow2x-5\sqrt{x}+2=0\)
\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=4\end{matrix}\right.\)
c) Ta có :
\(M=\frac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}=\frac{x+2\sqrt{x}+1}{\sqrt{x}}=\sqrt{x}+2+\frac{1}{\sqrt{x}}=2+\sqrt{x}+\frac{1}{\sqrt{x}}\)
AD - BDDT cô si cho 2 số nguyên dương \(\sqrt{x},\frac{1}{\sqrt{x}}\) ta có :
\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)
\(\Rightarrow M\ge2+2=4\)
Dấu = xảy ra khi \(\sqrt{x}=\frac{1}{\sqrt{x}}\Rightarrow x=1\)
Mà x ≠ 1 ⇒ M > 4