tách tử thành 1.3 ( cho 3 ra ngoài làm nhân tử chung)

=> ở mẫu còn nguyên tắc số thứ 2- số thứ 1 = tử

=> (1/1.2+1/2.3+.......+1/2015.2016 ) .3

 =  (2-1/1.2+3-2/2.3+......+2016-2015/2015.2016).3

 =  (2/1.2-1/1.2+3/2.3-2/2.3..........+2016/2015.2016- 2015/2015.2016).3

 =  ( 1-1/2+1/2-1/3+...........+ 1/2015-1/2016).3

 =   ( 1-1/2016 ) .3

 =    2015/2016 .3

\(S=3.\left(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}+...+\frac{1}{2015}.\frac{1}{2016}\right)\)

\(3S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}\)

\(3S=1-\frac{1}{2016}\)

\(3S=\frac{2015}{2016}\)

\(S=\frac{2015}{2016}:3\)

\(S=\frac{2015}{6048}\)

\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{2015.2016}\)

\(=3\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2016}\right)\)

\(=3\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)

\(=3\left(1-\frac{1}{2016}\right)\)

\(=3\cdot\frac{2015}{2016}=\frac{2015}{672}\)

sai bét

dễ lắm

s=

................

kb đi 

Ta có:

\(S=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+.......+\frac{3}{2015.2016}\)

\(\Rightarrow\frac{1}{3}.S=\frac{1}{1.2}+\frac{1}{2.3}+......+\frac{1}{2015.2016}\)

\(\Rightarrow\frac{1}{3}.S=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

\(\Rightarrow\frac{1}{3}.S=1-\frac{1}{2016}=\frac{2015}{2016}\)

\(\Rightarrow S=\frac{2015}{672}\)

Vậy: S = 2015/672

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2016\cdot2017}\)

\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2016}-\frac{1}{2017}\)

\(=1-\frac{1}{2017}=\frac{2016}{2017}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)

\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+......+\left(\frac{1}{2016}-\frac{1}{2017}\right)\)

\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2016}-\frac{1}{2017}\)

\(A=\frac{1}{1}-\frac{1}{2017}\)

\(A=\frac{2016}{2017}\)

A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{2016.2017}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2016}-\frac{1}{2017}\)

\(\Rightarrow A=1-\frac{1}{2017}\)

\(\Rightarrow A=\frac{2016}{2017}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2015.2016}+\frac{1}{2016.2017}\)

\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2015}-\frac{1}{2016}+\frac{1}{2016}-\frac{1}{2017}\)

\(\Rightarrow A=1-\frac{1}{2017}\)

\(\Rightarrow A=\frac{2016}{2017}\)

\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}=\frac{1}{5}\)

\(\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1^2.2^2.3^2.4^2}{1.2^2.3^2.4^2.5}=\frac{1}{5}\)

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2015.2106}\)

\(A=\left(\frac{1}{1}-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+...+\left(\frac{1}{2015}-\frac{1}{2016}\right)\)

\(A=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)

\(B=\frac{1}{2.4}+\frac{1}{4.6}+\frac{1}{6.8}+...+\frac{1}{2014.2016}=\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1007.1008}\right)\)

=> \(B=\frac{1}{4}.\left(\frac{1}{1}-\frac{1}{1008}\right)=\frac{1}{4}.\frac{1007}{1008}\)

=> \(B=\frac{1007}{4032}\)

\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+....+\frac{1}{1+2+3+...+2015}\)

\(=\frac{2}{1.2}+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+2+3\right).3}{2}}+.....+\frac{1}{\frac{\left(2015+1\right).2015}{2}}\)

\(=\frac{2}{1.2}+\frac{2}{2.3}+....+\frac{2}{2015.2016}\)

dễ vãi cả đạn

Thanks bạn nhiều tại câu này mk ko tk đúng đc cho bn nên mk tk đúng cho bn ở câu tl khác rồi nha

Chúc bn học tốt

hình như là 3² chứ k f 3³

\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)

\(B=\frac{\left(1\cdot1\right)\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)}{\left(1\cdot2\right)\left(2\cdot3\right)\left(3\cdot4\right)\left(4\cdot5\right)}\)

\(B=\frac{\left(1\cdot2\cdot3\cdot4\right)\left(1\cdot2\cdot3\cdot4\right)}{\left(1\cdot2\cdot3\cdot4\right)\left(2\cdot3\cdot4\cdot5\right)}\)

\(=\frac{1}{5}\)

\(B=\frac{1^2}{1\cdot2}\cdot\frac{2^2}{2\cdot3}\cdot\frac{3^2}{3\cdot4}\cdot\frac{4^2}{4\cdot5}\)

\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1\cdot2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5}\)

\(B=\frac{1^2\cdot2^2\cdot3^2\cdot4^2}{1^2\cdot2^2\cdot3^2\cdot4^2\cdot5}=\frac{1}{5}\)

Dii vào link nào nha : https://loga.vn/hoi-dap/tinh-tong-s-3-1-2-3-2-3-3-3-4-3-4-5-3-2015-20161-tinh-tong-s-dfrac-3-1-2-dfrac-3-2-3-dfrac-3-3-4-17371 

k mk