\(x=\frac{617}{1300}\)

\(x=\frac{617}{1300}\)

chúc học giỏi

trình bày ra dùm nha các bạn

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

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

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

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)=x-\frac{5}{13}\)

\(\frac{1}{3}-\left(\frac{1}{3}-\frac{1}{11}\right)=x-\frac{5}{13}\)

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

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

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

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

Giải:

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-...-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{1}{1.3}-\dfrac{1}{3.4}-\dfrac{1}{4.5}-...-\dfrac{1}{9.10}-\dfrac{1}{10.11}=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}\left(\dfrac{1}{1}-\dfrac{1}{11}\right)=x-\dfrac{5}{13}\)

\(\Leftrightarrow\dfrac{-1}{2}.\dfrac{10}{11}=x-\dfrac{5}{13}\)

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

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

\(\Leftrightarrow x=\dfrac{-10}{143}\)

Vậy ...

Giải đẳng cấp nhỉ @Hắc Hường

Ta có:

\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{5}{13}-\dfrac{1}{3}\\ -\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-...-\dfrac{1}{110}=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{10}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\left(\dfrac{1}{3}-\dfrac{1}{11}\right)=x-\dfrac{28}{39}\\ -\dfrac{8}{33}=x-\dfrac{28}{39}\\ x=-\dfrac{8}{33}+\dfrac{28}{39}\\ x=\dfrac{68}{143}\)

Vậy \(x=\dfrac{68}{143}\)

a)12,8 x 195 + 12,8 x 804 + 12,8

= 12,8 x (195+804+1)

=12,8 x 1000

=12800

a) = 12,8 \(\times\) ( 195 + 804 +1)

   = 12,8 \(\times\)  1000

   = 1280

b) = \(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{9\times10}\)

    = \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)

    = 1 - \(\frac{1}{10}\)

    = \(\frac{9}{10}\)

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

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

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

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

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

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

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

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

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

Em cảm ơn ạ.