bài 1:

\(\frac{6}{11}+\frac{1}{3}+\frac{5}{11}\)

\(=\left(\frac{6}{11}+\frac{5}{11}\right)+\frac{1}{3}\)

\(=\frac{11}{11}+\frac{1}{3}=1+\frac{1}{3}=\frac{3}{3}+\frac{1}{3}=\frac{4}{3}\)

bài 2:

\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}\)

\(=\left(\frac{1}{2}+\frac{1}{20}\right)+\left(\frac{1}{6}+\frac{1}{12}\right)\)

\(=\frac{11}{20}+\frac{1}{4}=\frac{11}{20}+\frac{5}{20}=\frac{15}{20}=\frac{3}{4}\)

bài 3: 

a) \(\frac{3}{2}\cdot\frac{4}{5}\cdot\frac{2}{3}=\left(\frac{3}{2}\cdot\frac{2}{3}\right)\cdot\frac{4}{5}=1\cdot\frac{4}{5}=\frac{4}{5}\)

b) \(\frac{6}{7}\cdot\frac{5}{3}\cdot\frac{7}{6}=\left(\frac{6}{7}\cdot\frac{7}{6}\right)\cdot\frac{5}{3}=1\cdot\frac{5}{3}=\frac{5}{3}\)

bài 4: 

a) \(\frac{2}{5}\cdot\frac{1}{4}+\frac{3}{4}\cdot\frac{2}{5}=\frac{2}{5}\cdot\left(\frac{1}{4}+\frac{3}{4}\right)=\frac{2}{5}\cdot1=\frac{2}{5}\)

b) \(\frac{6}{11}:\frac{2}{3}+\frac{5}{11}:\frac{2}{3}=\left(\frac{6}{11}+\frac{5}{11}\right):\frac{2}{3}=1:\frac{2}{3}=\frac{3}{2}\)

Bài 1: 

  6/11 + 1/3 + 5/11 

= ( 6/11 + 5/11) + 1/3

= 11/11 + 1/3

= 1 + 1/3 

= 3/3 +1/3  

= 4/3 

Bài 2: 

1/2 + 1/6 + 1/12 + 1/20 

= ( 1/2 + 1/6 + 1/12 ) + 1/20 

= ( 6/12 + 2/12 + 1/12 ) + 1/20 

=9/12 + 1/20 

= 3/4 +1/20  

= 15/20 + 1/20 

= 16/20 = 4/5

 Bài 3:

a) \(\frac{3}{2}\times\frac{4}{5}\times\frac{2}{3}\) \(=\left(\frac{3}{2}\times\frac{2}{3}\right)\times\frac{4}{5}\)\(=1\times\frac{4}{5}=\frac{4}{5}\) 

b)  \(\frac{6}{7}\times\left(\frac{5}{3}\times\frac{7}{6}\right)\) \(=\frac{6}{7}\times\frac{35}{18}\)\(=\frac{1\times5}{7\times3}=\frac{5}{21}\)   

Bài 4:

a) 2/5  x 1/4 + 3/4 x 2/5 

= 2/5 x ( 1/4  + 3/4) 

 = 2/5 x  1 

= 2/5

 b) 6/11 : 2/3 +5/11 : 2/3 

=  ( 6/11 + 5/11) x 3/2 

= 11/11 x 3/2  

= 1 x 3/2  

=  3/2  

....  

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{1\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}{4.\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4.\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)

\(=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left(\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right)}+\frac{3\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}{4\left(\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right)}\)

\(=\frac{1}{4}+\frac{3}{4}\)

=1

Ta có:

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}=\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{4\left[\frac{1}{9}-\frac{1}{7}-\frac{1}{11}\right]}=\frac{1}{4}\)

Lại có:

\(\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{3\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}{4\left[\frac{1}{5}-\frac{1}{25}-\frac{1}{125}-\frac{1}{625}\right]}=\frac{3}{4}\)

Vậy: 

\(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}=\frac{1}{4}+\frac{3}{4}=1\)

\(\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{5}{8}-\frac{5}{10}+\frac{5}{11}+\frac{5}{12}}+\frac{\frac{3}{2}+1+\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}+\frac{5}{4}}\)

\(=\frac{3.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}{5.\left(\frac{1}{8}-\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}{5.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}\)

\(=\frac{3}{5}+\frac{3}{5}\)

\(=\frac{6}{5}\)

1/6 nhe

\(=\frac{1}{6}\)

\(\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}.\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\cdot\frac{5^2}{5\cdot6}\)

\(=\frac{1\cdot1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot6}\)

\(=\frac{1}{6}\)

\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{5}{12}+1-\frac{7}{11}}\)

\(\Rightarrow P=\frac{\frac{8}{12}-\frac{3}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{11}{11}-\frac{7}{11}}\)

\(\Rightarrow P=\frac{\frac{5}{12}+\frac{5}{11}}{\frac{5}{12}+\frac{4}{11}}\)

Chúc bạn học tốt !!! 

mn giúp mk nha

mk đang cần gấp

\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{3}{12}+1+\frac{7}{11}}\)

\(P=\frac{\frac{2}{3}-\frac{1}{4}+\frac{5}{11}}{\frac{1}{4}+1+\frac{7}{11}}\)

\(P=\frac{\left(\frac{2}{3}-\frac{1}{4}+\frac{5}{11}\right)\cdot132}{\left(\frac{1}{4}+1+\frac{7}{11}\right)\cdot132}\)

\(P=\frac{88-33+60}{33+132+84}\)

\(P=\frac{115}{249}\)

Học nhiều quá quên mất bài dễ này rồi 

chuan

\(\frac{-7}{11}.\frac{11}{19}+\frac{-7}{11}.\frac{8}{19}+\frac{-4}{11}\)

\(=\frac{-7}{11}.\left(\frac{11}{19}+\frac{8}{19}\right)+\frac{-4}{11}\)

\(=\frac{-7}{11}.1+\frac{-4}{11}\)

\(=\frac{-7}{11}+\frac{-4}{11}=\frac{-11}{11}=-1\)

~ Hok tốt ~

Đặt \(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(\Rightarrow B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(\Rightarrow B=1-\frac{1}{2019}\)

\(\Rightarrow B=\frac{2018}{2019}\)

\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2018.2019}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2018}-\frac{1}{2019}\)

\(=1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{2018}-\frac{1}{2018}\right)-\frac{1}{2019}\)

\(=1-\frac{1}{2019}=\frac{2018}{2019}\)

~ Hok tốt ~