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

(5\(\dfrac{8}{17}\)+\(\dfrac{-4}{17}\)): x +\(\dfrac{22}{7}\):\(\dfrac{52}{3}\)=\(\dfrac{4}{11}\)

\(^{\dfrac{89}{17}}\):x+\(\dfrac{22}{7}\)x\(\dfrac{3}{52}\)=\(\dfrac{4}{11}\)

\(\dfrac{89}{17}\):x+\(\dfrac{33}{182}\)=\(\dfrac{4}{11}\)

\(\dfrac{89}{17}\):x=\(\dfrac{385}{2002}\)

x=\(\dfrac{2314}{85}\)

Cho link đây, hồi chiều mới làm: Câu hỏi của nguyen thu trang - Toán lớp 6 | Học trực tuyến

a: (x+1/2)(2/3-2x)=0

=>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

b: 

c: \(\Leftrightarrow x\cdot\left(\dfrac{13}{4}-\dfrac{7}{6}\right)=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{5}{12}+\dfrac{20}{12}=\dfrac{25}{12}\)

\(\Leftrightarrow x=\dfrac{25}{12}:\dfrac{39-14}{12}=\dfrac{25}{25}=1\)

a) 1/20 - (x - 8/5) = 1/10

x - 8/5 = 1/20 - 1/10

x - 8/5 = -1/20

x = -1/20 + 8/5

x = 31/20

b) 7/4 - (x + 5/3) = -12/5

x + 5/3 = 7/4 + 12/5

x + 5/3 = 83/20

x = 83/20 - 5/3

x = 149/60

c) x - [17/2 - (-3/7 + 5/3)] = -1/3

x - (17/2 - 26/21) = -1/3

x - 305/42 = -1/3

x = -1/3 + 305/42

x = 97/14

`a)4/5+5 1/2 xx (4,5-2)+7/10`

`=4/5+11/2*2,5+7/10`

`=0,8+2,2+0,7`

`=3+0,7=3,7`

`b)125%xx 17/4:(1 5/16-0,5)+2008`

`=1,25xx4,25:13/16+2008`

`=85/13+2008`

`=2014 7/13`

`c)5/11+(16/11+1)`

`=5/11+1+5/11+1`

`=2+10/11=32/11`

`d)3/17+11/4+5/8+14/17+3/8`

`=3/17+14/17+5/8+3/8+11/4`

`=1+1+11/4`

`=19/4`

a) 

\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)

<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)

<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)

b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)

= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)

c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)

= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)

d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)

= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)

a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)

\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)

\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)

\(\Leftrightarrow-7x=\dfrac{10}{3}\)

hay \(x=-\dfrac{10}{21}\)

b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)

\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)

\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)

\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)

\(\Leftrightarrow85x+85=84x+168\)

\(\Leftrightarrow x=83\)

\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(\dfrac{1}{3}.\left(1-\dfrac{1}{x+3}\right)=\dfrac{34}{103}\)

\(1-\dfrac{1}{x+3}=\dfrac{34}{103}:\dfrac{1}{3}=\dfrac{34}{103}.3\)

\(1-\dfrac{1}{x+3}=\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=1-\dfrac{102}{103}=\dfrac{103}{103}-\dfrac{102}{103}\)

\(\dfrac{1}{x+3}=\dfrac{1}{103}\)

\(\Rightarrow x+3=103\)

\(x=103-3\)

\(x=100\)

Vậy x = 100

\(5\dfrac{8}{17}:x+\left(\dfrac{-4}{17}\right):x+3\dfrac{1}{7}:17\dfrac{1}{3}=\dfrac{4}{11}\)

\(\Leftrightarrow\left(5\dfrac{8}{17}+\dfrac{-4}{17}\right):x+\dfrac{22}{7}:\dfrac{52}{3}=\dfrac{4}{11}\)

\(\Leftrightarrow\left(\dfrac{43}{17}+\dfrac{-4}{17}\right):x+\dfrac{22}{7}.\dfrac{3}{52}=\dfrac{4}{11}\)

\(\Leftrightarrow\dfrac{39}{17}:x+\dfrac{33}{182}=\dfrac{4}{11}\)

\(\Leftrightarrow\dfrac{39}{17}:x=\dfrac{4}{11}-\dfrac{33}{182}\)

(cứ thế mà làm tiếp nhé, mình thấy số xấu quá)