Question

[võ ngọc quốc uyên]

bài 2 thôi ạ

Answer 1

[Đoàn Đức Hà]

Bài 2. 

ĐKXĐ của biểu thức đã cho là: 

\(\hept{\begin{cases}x\ge0,\sqrt{x}\ne0\\\sqrt{x}-1\ne0\\\sqrt{x}-2\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>0\\x\ne1,x\ne2\end{cases}}\).

\(A=\left(\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}}\right)\div\left(\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)

\(=\frac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left[\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right]\)

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\div\left(\frac{x-1-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\right)\)

\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)=\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(A>\frac{1}{6}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}}>\frac{1}{6}\Leftrightarrow6\left(\sqrt{x}-2\right)>\sqrt{x}\)

\(\Leftrightarrow5\sqrt{x}>12\Leftrightarrow x>\frac{144}{25}\).