Question

1 TÌM x ϵ Z BIẾT:

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)≤ x ≤\(\frac{2}{3}\) .\(\left(|\frac{-1}{3}|-|\frac{-1}{2}|-|\frac{-3}{-4}|\right)\)

Answer 1

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{2}{3}.\left(\left|-\frac{1}{3}\right|-\left|-\frac{1}{2}\right|-\left|-\frac{3}{-4}\right|\right)\)

\(\Leftrightarrow-\frac{13}{3}.\frac{1}{3}\le x\le\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Leftrightarrow-\frac{13}{9}\le x\le\frac{2}{4}.-\frac{11}{12}\)

\(\Leftrightarrow-\frac{13}{9}\le x\le-\frac{11}{24}\)

\(\Rightarrow x\in\left\{-1,0\right\}\) ( do \(x\in Z\) )

Vậy : \(x\in\left\{-1,0\right\}\)

Answer 2

\(-4\frac{1}{3}\left(\frac{1}{2}-\frac{1}{6}\right)\le x\le\frac{2}{3}\left(|\frac{-1}{3}|-|\frac{-1}{2}|-|\frac{-3}{-4}|\right)\)

\(\Rightarrow\frac{-13}{9}\le x\le\frac{-11}{18}\)

\(\Rightarrow x\in\left[\frac{-13}{9};\frac{-11}{18}\right]\)