Giải:
\(\dfrac{-8}{15}< \dfrac{x}{40}\le\dfrac{-7}{15}\)
\(\Leftrightarrow\dfrac{-8.8}{15.8}< \dfrac{x.3}{40.3}\le\dfrac{-7.8}{15.8}\)
\(\Leftrightarrow\dfrac{-64}{120}< \dfrac{3x}{120}\le\dfrac{-56}{120}\)
\(\Leftrightarrow-64< 3x\le-56\)
\(\Leftrightarrow3x\in\left\{-63;-62;-61;-60;-59;-58;-57;-56\right\}\)
\(\Leftrightarrow x\in\left\{-21;-\dfrac{62}{3};-\dfrac{61}{3};-20;-\dfrac{59}{3};-\dfrac{58}{3};-19;-\dfrac{56}{3}\right\}\)
Mà x ∈ Z
\(\Leftrightarrow x\in\left\{-21;-20;-19\right\}\)
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