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

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

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

\(\Rightarrow x\in\){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 \(\left\{11;12;13;14\right\}\)

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

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

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

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

CHUC BAN HOC TOT :))

thiếu dấu ngoặc rồi thánh!

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

       \(\frac{13}{3}.\frac{-1}{3}\)   \(\le x\le\frac{2}{3}.\frac{-11}{12}\)

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

         \(\frac{-26}{18}\)     \(\le\frac{18x}{18}\le\frac{-11}{18}\)

Suy ra \(-26\le18x\le-11\)

\(\rightarrow x=-1\)

Vậy x = -1

\(\frac{3}{7}\cdot15\cdot\frac{1}{3}+\frac{3}{7}\cdot5\cdot\frac{2}{5}\le x\le\left(3\frac{1}{2}:7-6\frac{1}{2}\right)\cdot\left(-2\frac{1}{3}\right)\)

\(\Leftrightarrow\frac{15}{7}+\frac{6}{7}\le x\le-6\cdot\frac{-5}{3}\)

\(\Leftrightarrow3\le x\le10\)

Mà \(x\in Z\)

\(\Rightarrow x\in\left\{4;5;6;7;8;9\right\}\)

bn Quân sai rồi, hỗn số \(15\frac{1}{3}\)chứ có phải \(15.\frac{1}{3}\)đâu???

\(\frac{3}{7}\cdot15\frac{1}{3}+\frac{3}{7}\cdot5\frac{2}{5}\le x\le\left(3\frac{1}{2}:7-6\frac{1}{2}\right)\cdot\left(-2\frac{1}{3}\right)\)

\(\Rightarrow\frac{3}{7}\left(15\frac{1}{3}+5\frac{2}{5}\right)\le x\le\left(\frac{7}{2}:7-\frac{13}{2}\right)\cdot\left(-\frac{7}{3}\right)\)

\(\Rightarrow\frac{3}{7}\left(\frac{46}{3}+\frac{27}{5}\right)\le x\le\left(\frac{7}{14}-\frac{13}{2}\right)\cdot\left(-\frac{7}{3}\right)\)

\(\Rightarrow\frac{311}{35}\le x\le14\)

\(\Rightarrow8,8...\le x\le14\)

Đến đây là tìm x :))

-4/1/3.1/3< x <  -2/3.-11/12

-1/4/9< x < 11/18

-26/18< x < 11/18

Vậy x={-26/18;-25/18;.............;11/18}

khong biet

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

\(\Rightarrow-\frac{13}{3}.\left(\frac{3}{6}-\frac{1}{6}\right)\le x\le-\frac{2}{3}.\left(\frac{4}{12}-\frac{6}{12}-\frac{9}{12}\right)\)

\(\Rightarrow-\frac{13}{3}.\frac{2}{6}\le x\le-\frac{2}{3}.\frac{-11}{12}\)

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

\(\Rightarrow\frac{-26}{18}\le x\le\frac{11}{18}\)

=> -1,44444444444........... ≤ x ≤ 0,6111111111...........

Mà x ∈ Z

=> x ∈ { -1 ; 0 }

\(x\in\varnothing\)