a: \(\Leftrightarrow-\dfrac{23}{5}\cdot\dfrac{50}{23}< =x< =\dfrac{-13}{5}:\dfrac{21}{15}\)

=>-10<=x<=-13/7

hay \(x\in\left\{-10;-9;...;-2\right\}\)

b: \(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =-\dfrac{2}{3}\cdot\dfrac{-11}{12}\)

=>-13/9<=x<=11/18

hay \(x\in\left\{-1;0\right\}\)

`16/803+38+(-16/803)`
`=16/803-16/803+38`
`=0+38=38`

`100/91+310-9/91`
`=100/91-9/91+310`
`=1+310=311`

\(\dfrac{16}{103}+\left(38+\dfrac{-16}{103}\right)\)

\(=\dfrac{16}{103}+38+\dfrac{-16}{103}\)

\(=\dfrac{16}{103}+\dfrac{-16}{103}+38\)

\(=0+38\)

\(=38\)

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\)

\(\Rightarrow\dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\)

\(\Rightarrow-2\le x\le2\)

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

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

\(\Rightarrow\dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{6}{12}-\dfrac{2}{12}\right)\)

\(\Rightarrow\dfrac{2}{3}\cdot\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}\cdot\dfrac{4}{12}\)

\(\Rightarrow\dfrac{22}{36}\le\dfrac{x}{18}\le\dfrac{28}{36}\)

\(\Rightarrow\dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\)

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

8) \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}\left(\dfrac{6}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\le\dfrac{x}{18}\le\dfrac{7}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\\ \dfrac{2}{3}.\dfrac{11}{12}\le\dfrac{x}{18}\le\dfrac{7}{3}.\dfrac{2}{6}\\ \dfrac{11}{18}\le\dfrac{x}{18}\le\dfrac{14}{18}\\ \Rightarrow11\le x\le14\\ \Rightarrow x\in\left\{11;12;13;14\right\}\)

7) \(\dfrac{-5}{17}+\dfrac{3}{17}\le\dfrac{x}{17}\le\dfrac{13}{17}+\dfrac{-11}{17}\\ \dfrac{-2}{17}\le\dfrac{x}{17}\le\dfrac{2}{17}\\ \Rightarrow-2\le x\le2\\ \Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

\(\Leftrightarrow\dfrac{3}{7}\left(\dfrac{46}{3}+\dfrac{12}{5}\right)< =x< =\left(\dfrac{10}{3}:7-\dfrac{13}{2}\right)\cdot\dfrac{-7}{3}\)

\(\Leftrightarrow\dfrac{38}{5}< =x< =\dfrac{253}{18}\)

mà x là số nguyên

nên \(x\in\left\{8;9;10;11;12;13;14\right\}\)

c) \(\dfrac{x+1}{35}+\dfrac{x+2}{34}+\dfrac{x+3}{33}=\dfrac{x+4}{32}+\dfrac{x+5}{31}+\dfrac{x+6}{30}\)

\(\Rightarrow\dfrac{x+1}{35}+1+\dfrac{x+2}{34}+1+\dfrac{x+3}{33}+1=\dfrac{x+4}{32}+1+\dfrac{x+5}{31}+1+\dfrac{x+6}{30}+1\)

\(\Rightarrow\dfrac{x+1+35}{35}+\dfrac{x+2+34}{34}+\dfrac{x+3+33}{33}=\dfrac{x+4+32}{32}+\dfrac{x+5+31}{31}+\dfrac{x+6+30}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}=\dfrac{x+36}{32}+\dfrac{x+36}{31}+\dfrac{x+36}{30}\)

\(\Rightarrow\dfrac{x+36}{35}+\dfrac{x+36}{34}+\dfrac{x+36}{33}-\dfrac{x+36}{32}-\dfrac{x+36}{31}-\dfrac{x+36}{30}=0\)

\(\Rightarrow\left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\right)=0\)

\(\Rightarrow x+36=0\left(\text{vì }\dfrac{1}{35}+\dfrac{1}{34}+\dfrac{1}{33}+\dfrac{1}{32}+\dfrac{1}{31}+\dfrac{1}{30}\ne0\right)\)

\(\Rightarrow x=-36\)

Vậy ...

a/ Ta có: \(-4\dfrac{3}{5}.2\dfrac{4}{3}\le x\le-2\dfrac{3}{5}:1\dfrac{6}{15}\)

\(\Rightarrow\dfrac{-23}{5}.\dfrac{10}{3}\le x\le\dfrac{-13}{5}:\dfrac{21}{15}\)

\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{5}.\dfrac{15}{21}\)

\(\Rightarrow\dfrac{-46}{3}\le x\le\dfrac{-13}{7}\)

\(\Rightarrow-15,\left(3\right)\le x\le-1,\left(857142\right)\)

Vì x \(\in\) Z nên x \(\in\left\{-1;-2;-3;...;-15\right\}\)

Chúc bạn học tốt!!!

mk quên mất, còn câu b/ nữa, câu c thì bạn Nháy giải rồi nên mk khỏi giải nha.

b/\(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)\le x\le\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{7}\right)\)

\(\Rightarrow\)\(\dfrac{-13}{3}\left(\dfrac{3}{6}-\dfrac{1}{6}\right)\le x\le\dfrac{2}{3}\left(\dfrac{14}{42}-\dfrac{21}{42}-\dfrac{18}{42}\right)\)

\(\Rightarrow\dfrac{-13}{3}.\dfrac{1}{3}\le x\le\dfrac{2}{3}.\dfrac{-25}{42}\)

\(\Rightarrow\dfrac{-13}{9}\le x\le\dfrac{-25}{63}\)

\(\Rightarrow-1,\left(4\right)\le x\le0,\left(396825\right)\)

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

Mk chỉ nghĩ z thôi chứ có j thì góp ý thêm nha.

\(\Leftrightarrow-\dfrac{43}{5}\cdot\dfrac{90}{43}< =x< =\dfrac{-13}{5}:\dfrac{7}{5}\)

=>-18<=x<=-13/7

mà x là số nguyên

nên \(x\in\left\{-18;-17;...;-2\right\}\)

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

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

\(\dfrac{-13}{9}\le x\le\dfrac{-11}{12}\)

\(\dfrac{-468}{36}\le\dfrac{36.x}{36}\le\dfrac{-396}{36}\)

\(=>36.x\in\left\{-467;-466;-465;-464;...;-398;-397\right\}\)

\(=>x=-12\)

\(\Leftrightarrow-\dfrac{13}{3}\cdot\dfrac{1}{3}< =x< =\dfrac{-2}{3}\cdot\dfrac{4-6-9}{12}\)

\(\Leftrightarrow-\dfrac{13}{9}< =x< =\dfrac{-2}{3}\cdot\dfrac{-11}{12}=\dfrac{22}{36}=\dfrac{11}{18}\)

mà x là số nguyên

nên \(x\in\left\{-1;0;1\right\}\)