Ta có: \(A=\left(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\left(1-\frac{3-\sqrt{x}}{\sqrt{x}+1}\right)\) ( ĐKXĐ: \(x>0,\)\(x\ne0,\)\(x\ne1\))
\(\Leftrightarrow A=\left(\frac{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right).\left(x-\sqrt{x}+1\right)}{\sqrt{x}.\left(\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}+1-3+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}+1}\right)\)
\(\Leftrightarrow A=\left(\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow A=\left(\frac{2\sqrt{x}}{\sqrt{x}}\right).\left(\frac{\sqrt{x}+1}{2.\left(\sqrt{x}-1\right)}\right)\)
\(\Leftrightarrow A=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
Để \(A\ge\frac{3}{2}\)\(\Rightarrow\)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)
Ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1}\ge\frac{3}{2}\)
\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{3}{2}\ge0\)
\(\Leftrightarrow\frac{2\sqrt{x}+2-3\sqrt{x}+3}{2.\left(\sqrt{x}-1\right)}\ge0\)
\(\Leftrightarrow\frac{5-\sqrt{x}}{2.\left(\sqrt{x}-1\right)}\ge0\)
+ TH1: \(\hept{\begin{cases}5-\sqrt{x}\ge0\\2\sqrt{x}-2\ge0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\le5\\\sqrt{x}\ge1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\le25\\x\ge1\end{cases}}\)\(\Rightarrow\)\(1\le x\le25\)\(\left(TM\right)\)
+ TH2: \(\hept{\begin{cases}5-\sqrt{x}\le0\\2\sqrt{x}-2\le0\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}\sqrt{x}\ge5\\\sqrt{x}\le1\end{cases}}\)\(\Leftrightarrow\)\(\hept{\begin{cases}x\ge25\\x\le1\end{cases}}\)\(\left(L\right)\)
\(\Rightarrow\)\(1\le x\le25.\)Kết hợp ĐKXĐ: \(x\ne1\)
\(\Rightarrow\)\(1< x\le25\)
Vậy để \(A\ge\frac{3}{2}\)\(\Leftrightarrow\)\(1< x\le25\)
