\(x.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{78}\right)=\dfrac{220}{39}\)

\(x.\dfrac{20}{39}=\dfrac{220}{39}\)

\(x=\dfrac{220}{39}:\dfrac{20}{39}\)

x\(=11\)

⇔2x(13.4+14.5+..........+112.13)=22039

      \(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+\dfrac{x}{21}+\dfrac{x}{28}+\dfrac{x}{36}+\dfrac{x}{45}+\dfrac{x}{55}+\dfrac{x}{66}+\dfrac{x}{78}=\dfrac{220}{39}\)

⇔ \(x.\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+\dfrac{1}{45}+\dfrac{1}{55}+\dfrac{1}{66}+\dfrac{1}{78}\right)\) \(=\) \(\dfrac{220}{39}\)

⇔ \(x.\dfrac{20}{39}=\dfrac{220}{39}\)

⇔ \(x=11\)

\(\dfrac{x}{6}+\dfrac{x}{10}+\dfrac{x}{15}+........+\dfrac{x}{78}=\dfrac{220}{39}\)

\(\Leftrightarrow\dfrac{2x}{12}+\dfrac{2x}{20}+........+\dfrac{2x}{156}=\dfrac{220}{39}\)

\(\Leftrightarrow2x\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+..........+\dfrac{1}{12.13}\right)=\dfrac{220}{39}\)

\(\Leftrightarrow2x\left(\dfrac{1}{3}-\dfrac{1}{13}\right)=\dfrac{220}{39}\)

\(\Leftrightarrow2x.\dfrac{10}{39}=\dfrac{220}{39}\)

\(\Leftrightarrow x.\dfrac{20}{39}=\dfrac{220}{39}\)

\(\Leftrightarrow x=11\)

Vậy ...

Gọi biểu thức là A

\(A=\dfrac{2x}{12}+\dfrac{2x}{20}+\dfrac{2x}{30}+....+\dfrac{2x}{156}=\dfrac{200}{39}\)

Ta có công thức :

\(\dfrac{a}{b.c}=\dfrac{a}{c-b}.\left(\dfrac{1}{b}-\dfrac{1}{c}\right)\)

Áp dụng công thức trên, ta có :

\(A=\dfrac{2x}{3.4}+\dfrac{2x}{4.5}+\dfrac{2x}{5.6}+....+\dfrac{2x}{12.13}\)

\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+....+\dfrac{1}{12}-\dfrac{1}{13}\right)\)

\(A=2x.\left(\dfrac{1}{3}-\dfrac{1}{13}\right)\)

\(A=2x.\left(\dfrac{10}{39}\right)=\dfrac{200}{39}\)

\(A=2x=\dfrac{200}{39}:\dfrac{10}{39}\)

\(2x=20\)

\(\Rightarrow x=10\)

mink nghĩ vậy bạn ạ

\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)

\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)

\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

         \(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)

           \(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)

            \(x\) + 1 = 16

            \(x\)       = 16 - 1

             \(x\)     = 15 

a) Ta có: \(\dfrac{-11}{15}< \dfrac{x}{15}< \dfrac{-8}{15}\)

nên -11<x<-8

hay \(x\in\left\{-10;-9\right\}\)

b) Ta có: \(\dfrac{3}{7}< \dfrac{x}{21}< \dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{9}{21}< \dfrac{x}{21}< \dfrac{14}{21}\)

Suy ra: 9<x<14

hay \(x\in\left\{10;11;12;13\right\}\)

c) Ta có: \(\dfrac{-67}{21}< \dfrac{x}{168}< \dfrac{-3}{8}\)

nên \(\dfrac{-536}{168}< \dfrac{x}{168}< \dfrac{-63}{168}\)

Suy ra: -536<x<-63

hay \(x\in\left\{-535;-534;...;-64\right\}\)

\(\dfrac{1}{3.7}\)+\(\dfrac{1}{7.4}\) +\(\dfrac{1}{4.9}\) +...+\(\dfrac{2}{x\left(x+1\right)}\) =\(\dfrac{2}{9}\) 

\(\dfrac{2}{2.3.7}\)+\(\dfrac{2}{2.7.4}\) +\(\dfrac{2}{2.4.9}\) +...+\(\dfrac{2}{x\left(x+1\right)}\) =\(\dfrac{2}{9}\) 

\(\dfrac{2}{6.7}\)+\(\dfrac{2}{7.8}\) +\(\dfrac{2}{8.9}\) +...+\(\dfrac{2}{x\left(x+1\right)}\) =\(\dfrac{2}{9}\) 

2(\(\dfrac{1}{6.7}\) +\(\dfrac{1}{7.8}\) +\(\dfrac{1}{8.9}\) +...+\(\dfrac{1}{x\left(x+1\right)}\)) =\(\dfrac{2}{9}\) 

2(\(\dfrac{1}{6}\) -\(\dfrac{1}{7}\) +\(\dfrac{1}{7}\) -\(\dfrac{1}{8}\) +\(\dfrac{1}{8}\) -\(\dfrac{1}{9}\) +...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) ) =\(\dfrac{2}{9}\) 

2(\(\dfrac{1}{6}\) -\(\dfrac{1}{x+1}\) )=\(\dfrac{2}{9}\) 

\(\dfrac{1}{6}\)-\(\dfrac{1}{x+1}\) =\(\dfrac{2}{9}\) : 2

\(\dfrac{1}{6}\)-\(\dfrac{1}{x+1}\) =\(\dfrac{1}{9}\) 

\(\dfrac{1}{x+1}\) = \(\dfrac{1}{6}\) -\(\dfrac{1}{9}\) 

\(\dfrac{1}{x+1}\) = \(\dfrac{1}{18}\) 

x+1=18

x    = 18-1

x    =17

Vậy x =17

\(\dfrac{1}{21}+\dfrac{1}{28}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{6}-\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)

\(\Leftrightarrow x+1=18\)

\(\Leftrightarrow x=17\)

a)

\(\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{9}{38}\\ \dfrac{1}{4}-\dfrac{1}{x+3}=\dfrac{9}{38}\\\\ \dfrac{1}{x+3}=\dfrac{1}{4}-\dfrac{9}{38}\\ \dfrac{1}{x+3}=\dfrac{1}{76}\\ x+3=76\\ x=73.\)

b)

\(\dfrac{2}{42}+\dfrac{2}{56}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ \dfrac{2}{6.7}+\dfrac{2}{7.8}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\\ 2\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ 2.\left(\dfrac{1}{6}-\dfrac{1}{x+1}\right)=\dfrac{2}{9}\\ \dfrac{1}{x+1}=\dfrac{1}{6}-\dfrac{1}{9}=\dfrac{1}{18}\\ x+1=18\\ x=17.\)

e: =>2/7-x=2/5

=>7-x=5

=>x=2

f: =>2x+3/3=10/3

=>2x+3=10

=>2x=7

=>x=7/2

g: =>(14+x)/7=15/7

=>x+14=15

=>x=1

h: =>(2x+3)/x=13/x

=>2x+3=13

=>2x=10

=>x=5