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

giai tiếp nha bạn

Do \(\left|x-\dfrac{2}{3}\right|\ge0;\forall x\)

Mà \(-\dfrac{26}{\sqrt{81}}< 0\)

\(\Rightarrow\) Không tồn tại x để \(\left|x-\dfrac{2}{3}\right|< -\dfrac{26}{\sqrt{81}}\)

Hay ko tồn tại số nguyên x thỏa mãn đề bài

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

=> \(\left\{{}\begin{matrix}\frac{\sqrt{49}}{6}< \left|x-\frac{2}{3}\right|\\\left|x-\frac{2}{3}\right|< \frac{26}{\sqrt{81}}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\frac{7}{6}< \left|x-\frac{2}{3}\right|\\\left|x-\frac{2}{3}\right|< \frac{26}{9}\end{matrix}\right.\)

- TH₁ : \(x-\frac{2}{3}\ge0\left(x\ge\frac{2}{3}\right)\)

=> \(\left\{{}\begin{matrix}\frac{7}{6}< x-\frac{2}{3}\\x-\frac{2}{3}< \frac{26}{9}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\frac{11}{6}< x\\x< \frac{32}{9}\end{matrix}\right.\)

=> \(\frac{11}{6}< x< \frac{32}{9}\)

Mà x là số nguyên .

=> \(x\in\left\{2,3\right\}\)

- TH₂ : \(x-\frac{2}{3}< 0\left(x< \frac{2}{3}\right)\)

=> \(\left\{{}\begin{matrix}\frac{7}{6}< \frac{2}{3}-x\\\frac{2}{3}-x< \frac{26}{9}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\frac{1}{2}< -x\\-x< \frac{20}{9}\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}-\frac{1}{2}>x\\x>-\frac{20}{9}\end{matrix}\right.\)

=> \(-\frac{1}{2}>x>-\frac{20}{9}\)

Mà x là số nguyên .

=> \(x\in\left\{-1,-2\right\}\)

I là tham số à bạn

1. 

= -(13 + 3 căn7 ) / 2  +  -(7 + 3 căn7 ) / 2 

=  -7 + 3 căn7