\(\dfrac{1.2}{3.2}+\dfrac{1.2}{6.2}+.....+\dfrac{1.2}{2\left(x+1\right):2.2}\)=\(\dfrac{1998}{2000}\)

\(\dfrac{2}{6}+\dfrac{2}{12}+.....+\dfrac{2}{x\left(x+1\right)}\)=\(\dfrac{1998}{2000}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)

\(2\left(\dfrac{1}{2}-\dfrac{1}{x+1}\right)=\dfrac{1998}{2000}\)

\(\dfrac{1}{2}-\dfrac{1}{x+1}=\dfrac{1998}{2000}:2\)

\(\dfrac{1}{x+1}=\dfrac{1}{2}-\dfrac{999}{2000}\)

\(\dfrac{1}{x+1}=\dfrac{1}{2000}\)

suy ra x+1=2000

suy ra x=2000-1=1999

tick cho mk nha

1) \(\dfrac{5}{3}-x+\dfrac{1}{5}=\dfrac{-3}{10}\\ x+\dfrac{1}{5}=\dfrac{5}{3}-\left(\dfrac{-3}{10}\right)\\ x+\dfrac{1}{5}=\dfrac{5}{3}+\dfrac{3}{10}\\ x+\dfrac{1}{5}=\dfrac{59}{30}\\ x=\dfrac{59}{30}-\dfrac{1}{5}\\ x=\dfrac{53}{30}\)

2) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\\ \dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\\ \dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{8}{12}\\ \dfrac{2}{5}+x=\dfrac{1}{4}\\ x=\dfrac{1}{4}-\dfrac{2}{5}\\ x=\dfrac{-3}{20}\)

Sửa lại câu 1 nha :

\(\dfrac{5}{3}-x+\dfrac{1}{5}=\dfrac{-3}{10}\\ \dfrac{5}{3}-x=\dfrac{-3}{10}-\dfrac{1}{5}\\ \dfrac{5}{3}-x=\dfrac{-1}{2}\\ x=\dfrac{5}{3}-\left(\dfrac{-1}{2}\right)\\ x=\dfrac{13}{6}\)

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\\ \left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\\ \left(1-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\\ \dfrac{99}{100}-2x=\dfrac{1}{2}\\ 2x=\dfrac{99}{100}-\dfrac{1}{2}\\ 2x=\dfrac{49}{100}\\ x=\dfrac{49}{100}:2\\ x=\dfrac{49}{200}\)

\(\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)-2x=\dfrac{1}{2}\)

\([\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)]-2x=\dfrac{1}{2}\)

\(\left(\dfrac{1}{1}-\dfrac{1}{100}\right)-2x=\dfrac{1}{2}\)

\(\dfrac{99}{100}-2x=\dfrac{1}{2}\)

\(2x=\dfrac{99}{100}-\dfrac{1}{2}\)

\(2x=\dfrac{49}{100}\)

\(x=\dfrac{49}{100}:2\)

\(x=\dfrac{49}{200}\)

\(\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+\dfrac{1}{11\cdot14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\)\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)\(\Rightarrow x+3=308\Rightarrow x=305\)

\(\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{8.9.10}\right).x=\dfrac{23}{45}\)

\(\left[\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{8.9}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)\(\left[\dfrac{1}{2}.\left(\dfrac{1}{1.2}-\dfrac{1}{9.10}\right)\right].x=\dfrac{23}{45}\)

\(\left(\dfrac{1}{2}.\dfrac{22}{45}\right).x=\dfrac{23}{45}\)

\(\dfrac{11}{45}.x=\dfrac{23}{45}\)

\(x=\dfrac{23}{45}:\dfrac{11}{45}\)

\(x=\dfrac{23}{11}\)

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

\(\dfrac{x+24}{1996}+\dfrac{x+25}{1995}+\dfrac{x+26}{1994}+\dfrac{x+27}{1993}+\dfrac{x+2036}{4}=0\)

\(\Rightarrow\left(\dfrac{x+24}{1996}+1\right)+\left(\dfrac{x+25}{1995}+1\right)+\left(\dfrac{x+26}{1994}+1\right)+\left(\dfrac{x+27}{1993}+1\right)+\left(\dfrac{x+2036}{4}-4\right)=0\)\(\Rightarrow\dfrac{x+2020}{1996}+\dfrac{x+2020}{1995}+\dfrac{x+2020}{1994}+\dfrac{x+2020}{1993}+\dfrac{x+2020}{4}=0\)\(\Rightarrow\left(x+2020\right)\left(\dfrac{1}{9996}+\dfrac{1}{1995}+\dfrac{1}{1994}+\dfrac{1}{1993}+\dfrac{1}{4}\right)=0\)

\(\Rightarrow x+2020=0\Rightarrow x=-2020\)

1) \(2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\)

\(\Leftrightarrow2\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\)

\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\)

\(\Leftrightarrow\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=-\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{7}{8}+\dfrac{1}{3}\\\dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=-\dfrac{13}{24}\\\dfrac{1}{2}x=\dfrac{29}{24}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\left(-\dfrac{13}{24}\right):\dfrac{1}{2}\\x=\dfrac{29}{24}:\dfrac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{12}\\x=\dfrac{29}{12}\end{matrix}\right.\)

2) \(\dfrac{3}{4}-2\left|2x-\dfrac{2}{3}\right|=2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2\)

\(\Leftrightarrow2\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{8}:2\)

\(\Leftrightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{-5}{16}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{2}{3}=\dfrac{-5}{16}\\2x-\dfrac{2}{3}=\dfrac{5}{16}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-5}{16}+\dfrac{2}{3}\\2x=\dfrac{5}{16}+\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{17}{48}\\2x=\dfrac{47}{48}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{48}:2\\x=\dfrac{47}{48}:2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{17}{96}\\x=\dfrac{47}{96}\end{matrix}\right.\)

\(x+\dfrac{4}{5.9}+\dfrac{4}{9.13}+...+\dfrac{4}{41.45}=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{44}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\left(\dfrac{1}{5}-\dfrac{1}{45}\right)=-\dfrac{37}{45}\\ x+\dfrac{8}{45}=-\dfrac{37}{45}\\ x=-\dfrac{37}{45}-\dfrac{8}{45}\\ x=-1\)

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