\(\frac{2}{3}\) .\(\frac{3}{4}\)\(\le\)\(\frac{x}{18}\) \(\le\)\(\frac{7}{3}\).\(\frac{1}{3}\)

\(\frac{1}{2}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{9}{18}\le\frac{x}{18}\le\frac{14}{18}\)

\(\Rightarrow x\in\){9:10;11;12;13;14}

\(\frac{2}{3}.\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\)

\(\frac{2}{3}.\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}.\frac{1}{3}\)

\(\frac{2}{3}.\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)

Vậy \(x\in\left\{11;12;13\right\}\)

\(\frac{2}{3}\cdot\left(\frac{1}{2}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\left(\frac{1}{2}-\frac{1}{6}\right)\)

\(\frac{2}{3}\cdot\left(\frac{2}{4}+\frac{3}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\left(\frac{3}{6}-\frac{1}{6}\right)\)

\(\frac{2}{3}\cdot\left(\frac{5}{4}-\frac{1}{3}\right)\le\frac{x}{18}\le\frac{7}{3}\cdot\frac{1}{3}\)

\(\frac{2}{3}\cdot\left(\frac{15}{12}-\frac{4}{12}\right)\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{2}{3}\cdot\frac{11}{12}\le\frac{x}{18}\le\frac{7}{9}\)

\(\frac{11}{18}\le\frac{x}{18}\le\frac{14}{18}\)

Để \(x\)phải nhỏ hơn hoặc bằng thì x lần lượt bằng \(\left\{11;12;13;14\right\}\)

a)Ta có: 1/2-(1/3+1/4)= -1/12

           1/48-(1/16-1/6)=1/8

suy ra: -1/12<x<1/8

<=> -2/24<x<3/24

=>x thuộc:(-1/24 ;0 ;1/24 ;2/24 ;3/24)

x thuộc Z nhé các bạn

a)\(\frac{1}{2}-\left(\frac{1}{3}+\frac{1}{4}\right)< x< \frac{1}{48}-\left(\frac{1}{16}-\frac{1}{6}\right)\)

\(\Rightarrow\frac{1}{2}-\frac{7}{12}< x< \frac{1}{48}-\frac{-5}{48}\)

\(\Rightarrow\frac{-1}{12}< x< \frac{1}{8}\)

Mà \(x\in Z\Rightarrow x=0\)

b)\(4\frac{5}{9}:2\frac{5}{18}-7\le x< \frac{-2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow\frac{41}{9}:\frac{41}{18}-7\le x< \frac{-2}{3}.\frac{-11}{12}\)

\(\Rightarrow2-7\le x< \frac{11}{18}\)

\(\Rightarrow-5\le x< \frac{11}{18}\)

Mà \(x\in Z\Rightarrow x\in\){-5;-4;-3;-2;-1;0}

Chúc bạn học tốt

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

\(\Rightarrow\frac{13}{3}.\frac{1}{3}\le x\le\frac{2}{3}.\frac{-5}{12}\)

\(\Rightarrow\frac{13}{9}\le x\le\frac{-5}{18}\)

\(\Rightarrow\frac{26}{18}\le x\le\frac{-5}{18}\)

=> Không tìm được x

nhanh len nha. minh k cho

Sao đề khó đọc thế. Bạn viết rõ lại đi

\(a,4\frac{1}{3}\left[\frac{1}{2}-\frac{1}{6}\right]\le x\le-\frac{2}{3}\left[\frac{1}{3}\cdot\frac{1}{2}-\frac{3}{4}\right]\)

=> \(\frac{13}{3}\left[\frac{3}{6}-\frac{1}{6}\right]\le x\le-\frac{2}{3}\left[\frac{1}{6}-\frac{3}{4}\right]\)

=> \(\frac{13}{3}\cdot\frac{1}{3}\le x\le-\frac{2}{3}\cdot\left[\frac{2}{12}-\frac{9}{12}\right]\)

=> \(\frac{13}{9}\le x\le-\frac{2}{3}\cdot\left[-\frac{7}{12}\right]\)

=> \(\frac{13}{9}\le x\le-\frac{1}{3}\cdot\left[-\frac{7}{6}\right]\)

=> \(\frac{13}{9}\le x\le\frac{7}{18}\)

Đến đây tự tìm x 

\(\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}\le x\le\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)\)

\(taco:\left(\frac{-2}{3}-\frac{1}{2}\right):\frac{-1}{4}=\frac{-7}{6}:\frac{-1}{4}=\frac{14}{3}\)

\(\left(\frac{-5}{6}+\frac{2}{\frac{1}{4}}:\frac{-3}{2}\right)\cdot\left(\frac{-7}{\frac{1}{2}}\right)=\left(\frac{-5}{6}+\frac{-16}{3}\right)\cdot\left(-14\right)=\frac{-37}{6}\cdot\left(-14\right)=\frac{259}{3}\)

TU DO \(=>X=\frac{14}{3};\frac{15}{3};,,,;\frac{259}{3}\)

CHUC BAN HOC TOT :))

\(-\frac{13}{3}.\frac{1}{3}\le x\le-\frac{2}{3}.\left(-\frac{5}{12}\right)\)

\(-\frac{13}{9}\le x\le\frac{5}{18}\)

\(\frac{-26}{18}\le x\le\frac{5}{18}\)

\(x\in\left\{-\frac{26}{18};-\frac{25}{18};-\frac{24}{18};.....;\frac{4}{18};\frac{5}{18}\right\}\)