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

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

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

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

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

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

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

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

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

Em cảm ơn ạ.

a) \(2x-6=0\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=\dfrac{6}{2}=3\)

b) \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

còn câu c) d) nữa bạn ơi

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-5}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\\ \Leftrightarrow\dfrac{5\left(x+3\right)+4\left(x-3\right)}{x^2-9}=\dfrac{x-5}{x^2-9}\\ \Leftrightarrow5x+15+4x-12=x-5\\ \Leftrightarrow5x+4x-x=-5-15+12\\ \Leftrightarrow8x=-8\\ \Leftrightarrow x=-1\left(TM\right)\\ Vậy:S=\left\{-1\right\}\)

=>5(4x-1)-2+x<=3(10x-3)

=>20x-5+x-2<=30x-9

=>21x-7<=30x-9

=>-9x<=-2

=>x>=2/9

\(a,\left(\dfrac{31}{35}-\dfrac{4}{7}\right)\times\dfrac{8}{7}:2\\ =\left(\dfrac{31}{35}-\dfrac{4\times5}{35}\right)\times\dfrac{8}{7}:2\\ =\dfrac{11}{35}\times\dfrac{8}{7}:2\\ =\dfrac{88}{245}:2\\ =\dfrac{44}{245}\\ b,\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times\left(1-\dfrac{1}{5}\right)\\ =\left(\dfrac{2-1}{2}\right)\times\left(\dfrac{3-1}{3}\right)\times\left(\dfrac{4-1}{4}\right)\times\left(\dfrac{5-1}{5}\right)\\ =\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{3}\times\dfrac{3}{4}\times\dfrac{4}{5}\\ =\dfrac{1}{4}\times\dfrac{4}{5}=\dfrac{1}{5}\)

a, ( \(\dfrac{31}{35}\) - \(\dfrac{4}{7}\)) \(\times\) \(\dfrac{8}{7}\): 2

= \(\left(\dfrac{31}{35}-\dfrac{20}{35}\right)\) \(\times\) \(\dfrac{8}{7}\) : 2

= \(\dfrac{11}{35}\) \(\times\) \(\dfrac{8}{7}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{44}{35}\) \(\times\) \(\dfrac{4}{7}\)

= \(\dfrac{44}{245}\)

b, ( 1 - \(\dfrac{1}{2}\)) \(\times\) ( 1 - \(\dfrac{1}{3}\)) \(\times\) ( 1 - \(\dfrac{1}{4}\)) \(\times\) ( 1 - \(\dfrac{1}{5}\))

= \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2}{3}\) \(\times\) \(\dfrac{3}{4}\) \(\times\) \(\dfrac{4}{5}\)

= \(\dfrac{1}{5}\) \(\times\) \(\dfrac{2\times3\times4}{2\times3\times4}\)

= \(\dfrac{1}{5}\)

`[2-x]/x >= 1`

`<=>[2-x-x]/x >= 0`

`<=>[2-2x]/x >= 0`

`<=>0 < x <= 1`

     `->\bb B`

B

g: =>12x+1>=36x+12-24x-3

=>12x+1>=12x+9(loại)

h: =>6(x-1)+4(2-x)<=3(3x-3)

=>6x-6+8-4x<=9x-9

=>2x+2<=9x-9

=>-7x<=-11

=>x>=11/7

i: =>4x^2-12x+9>4x^2-3x

=>-12x+9>-3x

=>-9x>-9

=>x<1

\(\Leftrightarrow\dfrac{x-5}{95}-1+\dfrac{x-132}{32}+1=\dfrac{x-131}{31}+1+\dfrac{x-10}{90}-1\)

=>x-100=0

hay x=100