Question

tim x \(\in\) z biet

a . \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}\cdot\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\)

b. \(-4\dfrac{1}{3}\left(\dfrac{1}{2}+\dfrac{1}{6}\right)\le x\le\dfrac{-2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}\cdot\dfrac{3}{4}\right)\)

Answer 1

a. \(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{3}{4}\right)\le x\le\dfrac{1}{24}.\left(\dfrac{1}{3}-\dfrac{1}{3}\right)\)

\(\dfrac{1}{2}-\dfrac{13}{12}\le x\le\dfrac{1}{24}.0\) ( lười viết nên điền kết quả luôn )

\(\dfrac{-7}{12}\le x\le0\)

\(0,5833...\le x\le0\)

Vì \(x\in Z\)\(\Rightarrow x\in\left\{0\right\}\)

Vậy...

b. \(-4\dfrac{1}{3}\left(\dfrac{1}{2}+\dfrac{1}{6}\right)\le x\le\dfrac{-2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}.\dfrac{3}{4}\right)\)

\(\dfrac{-26}{9}\le x\le\dfrac{1}{36}\)

\(-2,8888...\le x\le0,277...\)

Vì \(x\in Z\Rightarrow x\in\left\{-2;-1;0\right\}\)

Vậy ...