\(a,5x\dfrac{7}{3}=\dfrac{5}{1}x\dfrac{7}{3}=\dfrac{35}{3};b,\dfrac{13}{4}:7=\dfrac{13}{4} :\dfrac{7}{1}=\dfrac{13}{4}x\dfrac{1}{7}=\dfrac{13}{28}\)

\(\dfrac{3}{7}+\dfrac{2}{5}+\dfrac{3}{4}=\dfrac{60}{140}+\dfrac{56}{140}+\dfrac{105}{140}=\dfrac{221}{140}\)

\(\dfrac{9}{7}-\dfrac{5}{11}x\dfrac{11}{7}=\dfrac{9}{7}-\dfrac{5}{7}=\dfrac{4}{7}\)

câu 1=19/12-7/12=12/12=1

a) Ta có : \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\\ x+2=0\Rightarrow x=-2\)

Lập bảng xét dấu:

TH : Xét x < -2

Ta có : - ( x+ 2) - (x - \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

-x - 2 -x + \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

- 2x - 2 + \(\dfrac{1}{2}\)= \(\dfrac{3}{4}\)

-2x = 2\(\dfrac{1}{4}\)

=> x = \(-1\dfrac{1}{8}\) ( loại )

TH 2: \(-2\le x< \dfrac{1}{2}\)

Ta có : x + 2 + ( -x + \(\dfrac{1}{2}\)) = \(\dfrac{3}{4}\)

=> \(2,5=\dfrac{3}{4}\) ( loại )

TH3 : \(x\ge\dfrac{1}{2}\)

x+ 2 + x - \(\dfrac{1}{2}\) = \(\dfrac{3}{4}\)

2x + 1,5 = \(\dfrac{3}{4}\)

x = -0,375( loại )

vậy ....

b) \(\left(\dfrac{2}{3}-2x\right).1\dfrac{1}{2}=\dfrac{3}{4}\\ \Rightarrow\dfrac{2}{3}-2x=-\dfrac{3}{4}\\ \Rightarrow2x=1\dfrac{5}{12}\\ \Rightarrow x=\dfrac{17}{24}\)

c) \(\left|x-1\right|+2.\left(x+4\right)=10\\ \Rightarrow\left|x-1\right|=10-2x-8\\ \Rightarrow\left|x-1\right|=2-2x\)

TH1 : \(x-1\ge0\) \(\Rightarrow x\ge1\)

\(\Rightarrow x-1=2-2x\\ \Rightarrow3x=3\\ \Rightarrow x=1\left(TM\right)\)

TH2 : \(x-1< 0\Rightarrow x< 1\)

=> \(x-1=-2+2x\\ \Rightarrow-x=-1\Rightarrow x=1\)(loại)

Vậy x = 1

b. \(\left(\dfrac{2}{3}-2x\right)\cdot1\dfrac{1}{2}=\dfrac{3}{4}\Rightarrow\dfrac{2}{3}-2x=\dfrac{3}{4}:\dfrac{3}{2}=\dfrac{1}{2}\Rightarrow-2x=\dfrac{1}{2}-\dfrac{2}{3}=-\dfrac{1}{6}\Rightarrow x=-\dfrac{1}{6}:\left(-2\right)=\dfrac{1}{12}\)

Vậy \(x=\dfrac{1}{12}\)

c. \(\left|x-1\right|+2\left(x+4\right)=10\Rightarrow\left|x-1\right|+2x+8=10\Rightarrow\left|x-1\right|+2x=10-8=2\)

\(\Rightarrow\left[{}\begin{matrix}x-1+2x=2;x-1\ge0\\-\left(x-1\right)+2x=2;x-1< 0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2x=2+1+3;x\ge1\\-x+1+2x=2;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=3;x\ge1\\-x+2x=2-1=1;x< 1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3:3=1;x\ge1\\x=1;x< 1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\end{matrix}\right.\)

Vậy x = 1

d. \(\dfrac{11}{12}+\dfrac{11}{12\cdot23}+...+\dfrac{11}{89\cdot100}+x=1\dfrac{2}{3}\)

\(\Rightarrow\dfrac{11}{12}+\dfrac{11}{276}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{22}{23}+...+\dfrac{11}{8900}+x=\dfrac{5}{3}\)

\(\Rightarrow\dfrac{99}{100}+x=\dfrac{5}{3}\Rightarrow x=\dfrac{5}{3}-\dfrac{99}{100}=\dfrac{203}{300}\)

Vậy \(x=\dfrac{203}{300}\)

e. \(\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x+4\dfrac{221}{231}=2\dfrac{1}{3}\)

\(\Rightarrow\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}\right)-x=\dfrac{7}{3}-4\dfrac{221}{231}\)

\(\Rightarrow x=\dfrac{\dfrac{7}{3}-\dfrac{1145}{231}}{\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{19\cdot21}}=\dfrac{-\dfrac{202}{77}}{\dfrac{2}{143}+\dfrac{2}{195}+...+\dfrac{2}{399}}=\dfrac{-\dfrac{202}{77}}{\dfrac{10}{231}}=\dfrac{-202}{77}\cdot\dfrac{231}{10}=\dfrac{-303}{5}\)

Vậy \(x=-\dfrac{303}{5}\)

\(a,\dfrac{5}{13}\times\dfrac{4}{15}\times13=\dfrac{5\times4\times13}{13\times5\times3}=\dfrac{4}{3}\\ b,\left(\dfrac{3}{7}+\dfrac{5}{2}\right)\times\dfrac{7}{5}=\dfrac{3}{7}\times\dfrac{7}{5}+\dfrac{5}{2}\times\dfrac{7}{5}=\dfrac{3}{5}+\dfrac{7}{2}=\dfrac{6}{10}+\dfrac{35}{10}=\dfrac{41}{10}\\ c,\dfrac{1}{5}\times\dfrac{11}{18}+\dfrac{11}{18}\times\dfrac{3}{5}=\dfrac{11}{18}\times\left(\dfrac{1}{5}+\dfrac{3}{5}\right)=\dfrac{11}{18}\times\dfrac{4}{5}=\dfrac{22}{45}\)

a, 4/3  b, 41/10   c,22/45

a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)

A=0+0+0+...+0+\(\dfrac{13}{15}\)

A=\(\dfrac{13}{15}\)

b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)

\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)

b, \(x+\dfrac{2}{3}\) = \(\dfrac{3}{5}\) - \(\dfrac{-1}{6}\)

    \(x+\dfrac{2}{3}\) = \(\dfrac{23}{30}\)

    \(x\)        = \(\dfrac{23}{30}\) - \(\dfrac{2}{3}\)

    \(x\)        = \(\dfrac{1}{10}\)

a, \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

                 \(\dfrac{2}{5}+x\)   = \(\dfrac{11}{12}-\dfrac{2}{3}\)

                 \(\dfrac{2}{5}-x\)   = \(\dfrac{1}{4}\)  

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

                          \(x=\dfrac{5}{12}\)

d, \(\dfrac{9}{2}\) - ( \(\dfrac{2}{3}\) - (\(x\) + \(\dfrac{7}{4}\))] = \(\dfrac{-5}{4}\)

     \(\dfrac{9}{2}\) - \(\dfrac{2}{3}\) + \(x\) + \(\dfrac{7}{4}\) = \(-\dfrac{5}{4}\)

      \(\dfrac{23}{6}\) + \(x\)            = - \(\dfrac{5}{4}\) - \(\dfrac{7}{4}\)

                \(x\)           = - 3  -  \(\dfrac{23}{6}\)

                \(x\)           = - \(\dfrac{41}{6}\)

a) \(\dfrac{11}{10}+\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{11}{10}+\dfrac{3}{5}\times\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)

b) \(\dfrac{4}{3}+5\times\dfrac{5}{8}=\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)

c) \(\left(\dfrac{2}{5}+\dfrac{3}{7}\right)\times\dfrac{25}{29}=\left(\dfrac{14}{35}+\dfrac{15}{35}\right)\times\dfrac{25}{39}=\dfrac{29}{35}\times\dfrac{25}{39}=\dfrac{145}{274}\)

d) \(\dfrac{1}{4}\times\dfrac{5}{12}+\dfrac{5}{12}\times\dfrac{4}{5}=\dfrac{5}{12}\times\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}\times\dfrac{21}{20}=\dfrac{105}{240}=\dfrac{7}{16}\)

a) \(\dfrac{11}{10}+\dfrac{3}{5}x\dfrac{3}{2}=\dfrac{11}{10}+\dfrac{9}{10}=\dfrac{20}{10}=2\)

b) \(\dfrac{4}{3}+\dfrac{25}{8}=\dfrac{32}{24}+\dfrac{75}{24}=\dfrac{107}{24}\)

c) \(\dfrac{29}{35}x\dfrac{25}{29}=\dfrac{5}{7}\)

\(=\dfrac{5}{12}x\left(\dfrac{1}{4}+\dfrac{4}{5}\right)=\dfrac{5}{12}x\dfrac{21}{20}=\dfrac{7}{16}\)

a: \(=\dfrac{-6}{11}:\dfrac{3\cdot11}{4\cdot5}=\dfrac{-6}{11}\cdot\dfrac{20}{33}=\dfrac{-2}{11}\cdot\dfrac{20}{11}=\dfrac{-40}{121}\)

b: \(=\dfrac{7}{12}+\dfrac{5}{72}-\dfrac{11}{36}=\dfrac{42}{72}+\dfrac{5}{72}-\dfrac{22}{72}=\dfrac{25}{72}\)

c: \(=\dfrac{13}{10}:\dfrac{-5}{13}=\dfrac{-169}{50}\)