Question

tìm x biết \(\dfrac{-15}{41}+\dfrac{-138}{41}\le x\le\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{6}\)

Answer 1

\(\Leftrightarrow\dfrac{-153}{41}< =x< =1\)

mà x là số nguyên

nên \(x\in\left\{-3;-2;-1;0;1\right\}\)