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\(\dfrac{x-1}{12}+\dfrac{x-1}{20}+\dfrac{x-1}{30}+...+\dfrac{x-1}{72}=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{72}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{8.9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\\ \left(x-1\right)\dfrac{2}{9}=\dfrac{16}{9}\\ x-1=8\\ x=8+1\\ x=9\)

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)

\(\dfrac{1}{3}\)  - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)

 \(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) =  \(x\) - \(\dfrac{5}{13}\)

\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) +  \(\dfrac{1}{11}\) =  \(x\) - \(\dfrac{5}{13}\)

         \(x-\dfrac{5}{13}=\dfrac{1}{11}\)

        \(x\)           = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)

      \(x\)           = \(\dfrac{68}{143}\)

Em cảm ơn ạ.

\(2\dfrac{2}{9}-x=\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)

\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{7.8}+\dfrac{1}{8.9}\)

\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\)

\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{1}{3}-\dfrac{1}{9}\)

\(\Rightarrow2\dfrac{2}{9}-x=\dfrac{2}{9}\)

\(\Rightarrow x=2\dfrac{2}{9}-\dfrac{2}{9}\)

\(\Rightarrow x=2\)

Vậy x=2

2\(\dfrac{2}{9}\) - x = \(\dfrac{1}{3\cdot4}\)+\(\dfrac{1}{4\cdot5}\)+\(\dfrac{1}{5\cdot6}\)+\(\dfrac{1}{6\cdot7}\)+\(\dfrac{1}{7\cdot8}\)+\(\dfrac{1}{8\cdot9}\)

2\(\dfrac{2}{9}\)-x = \(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)+...+\(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)

2\(\dfrac{2}{9}\)-x = \(\dfrac{1}{3}\)-\(\dfrac{1}{9}\)

2\(\dfrac{2}{9}\)-x = \(\dfrac{9}{27}\)- \(\dfrac{3}{27}\)

2\(\dfrac{2}{9}\)-x = \(\dfrac{2}{9}\)

\(\dfrac{20}{9}\) -x = \(\dfrac{2}{9}\)

x = \(\dfrac{20}{9}-\dfrac{2}{9}\)

x = 2

Vậy x = 2

a: \(\Leftrightarrow\dfrac{32}{x}=\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{99}\)

=>32/x=1/3-1/5+1/5-1/7+...+1/9-1/11

=>32/x=1/3-1/11=8/33

=>x=32:8/33=132

b: \(\Leftrightarrow1-\dfrac{1}{6}+1-\dfrac{1}{12}+...+1-\dfrac{1}{56}=\dfrac{x}{16}\)
\(\Leftrightarrow6-\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\right)=\dfrac{x}{16}\)

=>x/16=6-1/2+1/8=11/2+1/8=45/8=90/16

=>x=90

c: \(\Leftrightarrow\dfrac{22}{x}=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{10}\right)\left(1+\dfrac{1}{10}\right)\)

=>22/x=1/2*2/3*...*9/10*3/2*4/3*...*11/10

=>22/x=1/10*11/2=11/20=22/40

=>x=40

Ta có:

\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)

\(\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{4-3}{3.4}+\dfrac{5-4}{4.5}+\dfrac{6-5}{5.6}+\dfrac{7-6}{6.7}+\dfrac{8-7}{7.8}+\dfrac{9-8}{8.9}+\dfrac{10-9}{9.10}\right)-x=\dfrac{-19}{24}\)

\(\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)\(\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)

\(\dfrac{7}{30}-x=\dfrac{-19}{24}\)

\(x=\dfrac{7}{30}-\dfrac{-19}{24}\)

\(x=\dfrac{41}{40}\)

\(\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\left(\dfrac{1}{3}-\dfrac{1}{10}\right)-x=\dfrac{-19}{24}\)

\(\Leftrightarrow\dfrac{7}{30}-x=\dfrac{-19}{24}\)

\(\Rightarrow x=\dfrac{7}{30}-\dfrac{-19}{24}\)

\(\Rightarrow x=\dfrac{41}{40}\)

( 1/3.4 +1/4.5+ 1/5.6+ 1/6.7+1/7.8+1/8.9+1/9.10)-x=-19/24

(1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)-x=-19/24

(1/3-1/10)-x=-19/24

7/30-x=-19/24

x=7/30-(-19)/24

x=41/40

nếu dung tick cho mình nhá

a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)

\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)

=>2007-x=0

hay x=2007

b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)

\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)

=>x+7+1/3-1/10=0

hay x=-217/30

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

\(\Leftrightarrow\left(x+1\right)^2=4\)

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

b: Ta có: \(\dfrac{\left(x-2\right)^2}{7}=\dfrac{49}{\left(x-2\right)}\)

\(\Leftrightarrow x-2=7\)

hay x=9

d: ĐKXĐ: x<>-4; x<>-5; x<>-6; x<>-7

\(PT\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)

=>\(\dfrac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)

=>x^2+11x+28=54

=>x^2+11x-26=0

=>(x+13)(x-2)=0

=>x=2 hoặc x=-13

e: \(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Leftrightarrow\left(\dfrac{x-241}{17}-1\right)+\left(\dfrac{x-220}{19}-2\right)+\left(\dfrac{x-195}{21}-3\right)+\left(\dfrac{x-166}{23}-4\right)=0\)

=>x-258=0

=>x=258

a) Ta có: \(A=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2014}\right)\left(1-\dfrac{1}{2015}\right)\left(1-\dfrac{1}{2016}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2013}{2014}\cdot\dfrac{2014}{2015}\cdot\dfrac{2015}{2016}\)

\(=\dfrac{1}{2016}\)

b) Ta có: \(\dfrac{x-2}{12}+\dfrac{x-2}{20}+\dfrac{x-2}{30}+\dfrac{x-2}{42}+\dfrac{x-2}{56}+\dfrac{x-2}{72}=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\left(\dfrac{1}{3}-\dfrac{1}{9}\right)=\dfrac{16}{9}\)

\(\Leftrightarrow\left(x-2\right)\cdot\dfrac{2}{9}=\dfrac{16}{9}\)

\(\Leftrightarrow x-2=\dfrac{16}{9}:\dfrac{2}{9}=\dfrac{16}{9}\cdot\dfrac{9}{2}=8\)

hay x=10

Vậy: x=10