Question

Tìm số nguyên x biết

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

Answer 1

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

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

Rùi tự thay vào 

Answer 2

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

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

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

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

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

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

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