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a) Ta có:
\(-\dfrac{3}{4}< \dfrac{a}{12}< -\dfrac{5}{9}\)
hay \(-\dfrac{27}{36}< \dfrac{3a}{36}< -\dfrac{20}{36}\)
\(\Rightarrow-27< 3a< -20\)
\(\Rightarrow3a\in\left\{-26;-25;-24;-23;-22;-21\right\}\)
Mà \(a\) là số nguyên suy ra:
\(a=-8\)
a) \(\frac{2}{-5}< \frac{x}{10}< \frac{1}{4}\)
\(\Rightarrow\frac{-8}{20}< \frac{2x}{20}< \frac{5}{20}\)
\(\Rightarrow-8< 2x< 5\)
\(\Rightarrow-4< x< 2,5\)
Vì \(x\inℤ\) nên \(x\in\left\{-3;-2;-1;0;1;2\right\}\)
b) \(-\frac{2}{3}< \frac{x}{8}< -\frac{1}{6}\)
\(\Rightarrow\frac{-16}{24}< \frac{3x}{24}< \frac{-4}{24}\)
\(\Rightarrow-16< 3x< -4\)
\(\Rightarrow3x\in\left\{-15;-12;-9;-6\right\}\)
\(\Rightarrow x\in\left\{-5;-4;-3;-2\right\}\)
a) \(\dfrac{7}{4}< \dfrac{a}{8}< 3\\ =>\dfrac{7}{4}.8< a< 3.8\\ =>14< a< 24\\ =>a\in\left\{15;16;17;...;23\right\}\)
b) \(\dfrac{2}{3}< \dfrac{a-1}{6}< \dfrac{8}{9}\\ =>\dfrac{2}{3}.6< a-1< \dfrac{8}{9}.6\\ =>4< a-1< \dfrac{16}{3}\\ =>4+1< a< \dfrac{16}{3}+1\\ =>5< a< \dfrac{19}{3}\\ =>a=6\)
b) \(\dfrac{2}{3}< a-\dfrac{1}{6}< \dfrac{8}{9}\\ =>\dfrac{2}{3}+\dfrac{1}{6}< a< \dfrac{8}{9}+\dfrac{1}{6}\\ =>\dfrac{5}{6}< a< \dfrac{19}{18}\\ =>a=1\)
c) \(\dfrac{12}{9}< \dfrac{4}{a}< \dfrac{8}{3}\\ =>\dfrac{24}{18}< \dfrac{24}{6a}< \dfrac{24}{9}\\ =>9< 6a< 18\\ =>\dfrac{9}{6}< a< \dfrac{18}{6}\\ =>1,5< a< 3\\ =>a=2\)