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\(a,\frac{1}{3}+\frac{6}{11}.\frac{22}{5}-\frac{3}{4}:1,5=\frac{1}{3}+\frac{6.2}{5}-\frac{3}{4}:\frac{3}{2}=\frac{1}{3}+\frac{12}{5}-\frac{1}{2}=\frac{10}{30}+\frac{72}{5}-\frac{15}{30}=\frac{67}{30}\)

\(b, 3.3,5.6+2.6,4.9+1,8=18.3,5+18.6,4+1,8=18\left(3,5+6,4\right)+1,8=18.9,9+1,8=1,8.99+1,8=1,8\left(1+99\right)=1,8.100=180\)

b ) 3 x 3,5 x 6 + 2 x 6,4 x 9 + 1,8 

= 18 x 3,5 + 18x 6,4 + 18 x 0,1 

= 18 ( 3,5 + 6,4 + 0,1)

= 18 x 10 

= 180 

a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)

\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)

=>2x+10=0

hay x=-5

b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)

=>x-2017=0

hay x=2017

a: Sửa đề: \(\dfrac{2x-1}{11}+\dfrac{2x-2}{12}+\dfrac{2x-3}{13}=\dfrac{2x+5}{5}+\dfrac{2x+7}{3}+\dfrac{2x+4}{6}\)

\(\Leftrightarrow\dfrac{2x-1}{11}+1+\dfrac{2x-2}{12}+1+\dfrac{2x-3}{13}+1=\dfrac{2x+5}{5}+1+\dfrac{2x+7}{3}+1+\dfrac{2x+4}{6}+1\)

=>2x+10=0

hay x=-5

b: \(\dfrac{x-1}{2016}+\dfrac{x-2}{2015}+\dfrac{x-3}{2014}+\dfrac{x-4}{2013}+\dfrac{x-5}{2012}-5=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{2016}-1\right)+\left(\dfrac{x-2}{2015}-1\right)+\left(\dfrac{x-3}{2014}-1\right)+\left(\dfrac{x-4}{2013}-1\right)+\left(\dfrac{x-5}{2012}-1\right)=0\)

=>x-2017=0

hay x=2017