\(\Rightarrow\frac{7}{6}< |x-\frac{2}{3}|< \frac{26}{9}\)

\(\Rightarrow\frac{21}{18}< |x-\frac{2}{3}|< \frac{52}{18}\)

Rùi tự thay vào 

\(\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\)

\(\Leftrightarrow\frac{7}{6}< \left|x-\frac{2}{3}\right|< \frac{26}{9}\)

\(\Leftrightarrow\frac{7}{6}< 2\le\left|x-\frac{2}{3}\right|\le2< \frac{26}{9}\)

\(\Leftrightarrow\left|x-\frac{2}{3}\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=2\\x-\frac{2}{3}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=--\frac{4}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{8}{3};-\frac{4}{3}\right\}\)

\(\frac{7}{6}< |a-\frac{2}{3}|< \frac{26}{9}\Leftrightarrow\frac{21}{18}< |a-\frac{2}{3}|< \frac{52}{18}\Leftrightarrow\orbr{\begin{cases}-\frac{52}{18}< a-\frac{2}{3}< -\frac{21}{18}\\\frac{21}{18}< a-\frac{2}{3}< \frac{52}{18}\end{cases}}\) 

giai tiếp nha bạn

ĐKCĐ: \(x\ge0;x\ne9,x\ne4\)

\(A=\left(\frac{x-3\sqrt{x}}{x-9}-1\right):\left(\frac{9-x}{x+\sqrt{x}-6}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\\ \)

   \(=\left(\frac{\sqrt{x}.\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}-1\right):\left(\frac{\left(3-\sqrt{x}\right).\left(3+\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x+3}\right)}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

  \(=\left(\frac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\frac{3-\sqrt{x}}{\sqrt{x}-2}+\frac{\sqrt{x}-3}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)

   \(=-\frac{3}{\sqrt{x}+3}:\left(-\frac{\sqrt{x}-2}{\sqrt{x}+3}\right)=-\frac{3}{\sqrt{x}+3}:\frac{-\left(\sqrt{x}-2\right)}{\sqrt{x}+3}=\frac{3}{\sqrt{x}-2}\)

b, \(A\inℤ\Leftrightarrow\frac{3}{\sqrt{x}-2}\inℤ\)

Nếu x không là số chính phương thì  \(\sqrt{x}\)là số vô tỉ thì \(\sqrt{x}-2\)là số vô tỉ\(\Rightarrow A=\frac{3}{\sqrt{x}-2}\)là số vô tỉ

Nếu x là số chính phương thì \(\sqrt{x}\)là số nguyên thì \(\sqrt{x}-2\inℤ\Rightarrow\sqrt{x}-2\inƯ\left(3\right)\Rightarrow\sqrt{x}-2\in\left\{\pm1;\pm3\right\}\Rightarrow\sqrt{x}\in\left\{1;3;5\right\}\)\(\Rightarrow x\in\left\{1;9;25\right\}\)

Mà theo ĐKXĐ có x khác 9 => \(x\in\left\{1,25\right\}\)