a, \(\left|x-\dfrac{5}{3}\right|< \dfrac{1}{3}\)
⇒ \(-\dfrac{1}{3}\) < \(x-\dfrac{5}{3}\)< \(\dfrac{1}{3}\)
⇒ \(-\dfrac{1}{3}+\dfrac{3}{5}\) < x < \(\dfrac{1}{3}+\dfrac{3}{5}\)
⇒ \(\dfrac{4}{15}\) < x < \(\dfrac{14}{15}\)
Vậy\(\dfrac{4}{15}\)< x < \(\dfrac{14}{15}\)
b,\(\left|x+\dfrac{11}{2}\right|\)> |−5,5|
⇒ \(\left|x+\dfrac{11}{2}\right|\)> 5,5
⇒ \(\left[{}\begin{matrix}x+\dfrac{11}{2}\ge5,5\\x+\dfrac{11}{2}\le-5,5\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x\ge0\\x\le11\end{matrix}\right.\)
Vậy 0 ≤ x ≤ 11
c, \(\dfrac{2}{5}< \left|x-\dfrac{7}{5}\right|< \dfrac{3}{5}\)
⇒ \(\pm\dfrac{2}{5}< x-\dfrac{7}{5}< \pm\dfrac{3}{5}\)
Xét \(\dfrac{2}{5}< x-\dfrac{7}{5}< \dfrac{3}{5}\)
⇒\(\dfrac{2}{5}+\dfrac{7}{5}< x< \dfrac{3}{5}+\dfrac{7}{5}\)
⇒ \(\dfrac{9}{5}< x< 2\)
Xét \(-\dfrac{2}{5}>x-\dfrac{7}{5}>-\dfrac{3}{5}\)
⇒ \(-\dfrac{2}{5}+\dfrac{7}{5}>x>-\dfrac{3}{5}+\dfrac{7}{5}\)
⇒ \(1>x>\dfrac{4}{5}\)
Vậy \(\left[{}\begin{matrix}\dfrac{9}{5}< x< 2\\\dfrac{4}{5}< x< 1\end{matrix}\right.\)
Tick mk nhé ☺
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