a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)
nên -11<x<-8
hay \(x\in\left\{-10;-9\right\}\)
b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)
Suy ra: 9<x<14
hay \(x\in\left\{10;11;12;13\right\}\)
c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)
nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)
Suy ra: -536<x<-63
hay \(x\in\left\{-535;-534;...;-64\right\}\)