\(=\dfrac{1}{2}\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{13\cdot15}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{4}{15}=\dfrac{2}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)

\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)

1100444-88888=

a)\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}\)

\(=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right)\)

\(=\frac{1}{2}.\left(\frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{11}\right)\)

\(\frac{10}{22}\)

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}=\dfrac{10}{11}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)

\(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\)

\(=2\left(\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}\right)\)

\(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}\)

\(=\dfrac{1}{3}-\dfrac{1}{13}\)

\(=\dfrac{10}{39}\)

nhưng khi tính trong máy tính được kết quả là \(\dfrac{5}{39}\) mà bạn

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{99.101}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)

đúng 

`a)1/2 . [-3]/4 . [-5]/8 . [-8]/9=[1. (-3).(-5).(-8)]/[2.4.8.3.3]=[-5]/[2.4.3]=[-5]/24`

`b)(2/[1.3]+2/[3.5]+2/[5.7]).([10.13]/3-[2^2]/3-[5^3]/3)`

`=(1-1/3+1/3-1/5+1/5-1/7).[10.13-2^2-5^3]/3`

`=(1-1/7).[130-4-125]/3`

`=6/7 . 1/3 = 2/7`

____________________________________________________

`8/9+1/9 . 2/9+1/9 . 7/9`

`=8/9+1/9.(2/9+7/9)`

`=8/9+1/9 . 9/9`

`=8/9+1/9=9/9=1`

a) ^{_{\(\dfrac{1}{2}\cdot\dfrac{-3}{4}\cdot\dfrac{-5}{8}\cdot\dfrac{-8}{9}\)}}

^{_{\(=\dfrac{1\cdot\left(-3\right)\cdot\left(-5\right)\cdot\left(-8\right)}{2\cdot4\cdot8\cdot9}\)}}

^{_{\(=-\dfrac{5}{24}\)}}

b) ^{_{\(\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}\right)\cdot\left(\dfrac{10\cdot13}{3}-\dfrac{2^2}{3}-\dfrac{5^3}{3}\right)\)}}

^{_{\(=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}\right)\cdot\left(\dfrac{130}{3}-\dfrac{4}{3}-\dfrac{125}{3}\right)\)}}

_{^{\(=\left(1-\dfrac{1}{7}\right)\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{6}{7}\cdot\dfrac{1}{3}\)}}

_{^{\(=\dfrac{2}{7}\)}}

_{^{\(\dfrac{8}{9}+\dfrac{1}{9}\cdot\dfrac{2}{9}+\dfrac{1}{9}\cdot\dfrac{7}{9}\)}}

_{^{\(=\dfrac{8}{9}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_{^{\(=\dfrac{72}{81}+\dfrac{2}{81}+\dfrac{7}{81}\)}}

_^\(=1\)

\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)

\(\Rightarrow\dfrac{M}{2}=\dfrac{6:2}{2.5}+...+\dfrac{6:2}{47.50}\)

\(=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{47.50}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{47}-\dfrac{1}{50}\)

\(=\dfrac{1}{2}-\dfrac{1}{50}\)

\(=\dfrac{12}{25}\)

\(\Rightarrow M=\dfrac{12}{25}.2=\dfrac{24}{25}\)

\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)

\(\Rightarrow2K=\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{43.45}\)

\(=\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\)

\(=\dfrac{1}{9}-\dfrac{1}{45}\)

\(=\dfrac{4}{45}\)

\(\Rightarrow K=\dfrac{4}{45}:2=\dfrac{2}{45}\)

\(M=\dfrac{6}{2.5}+\dfrac{6}{5.8}+\dfrac{6}{8.11}+...+\dfrac{6}{47.50}\)

\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{5}+\dfrac{6}{5}-\dfrac{6}{8}+\dfrac{6}{8}-\dfrac{6}{11}+...+\dfrac{6}{47}-\dfrac{6}{50}\right)\)

\(M=\dfrac{6}{3}.\left(\dfrac{6}{2}-\dfrac{6}{50}\right)\)

\(M=\dfrac{6}{3}.\left(\dfrac{150}{50}-\dfrac{6}{50}\right)\)

\(M=\dfrac{6}{3}.\dfrac{144}{50}\)

\(M=\dfrac{144}{25}\)

\(K=\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+...+\dfrac{1}{43.45}\)

\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\right)\)

\(K=\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{45}\right)\)

\(K=\dfrac{1}{2}.\left(\dfrac{5}{45}-\dfrac{1}{45}\right)\)

\(K=\dfrac{1}{2}.\dfrac{4}{45}\)

\(K=\dfrac{2}{45}\)

a, A= \(5\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\)

\(A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=5\left(1-\dfrac{1}{100}\right)\)

\(A=5.\dfrac{99}{100}=\dfrac{99}{20}.\)

b, \(C=1.2.3+2.3.4+...+8.9.10\)

\(4C=1.2.3.4+2.3.4.\left(5-1\right)+...+8.9.10.\left(11-7\right)\)\(4C=1.2.3.4+2.3.4.5-1.2.3.4+...+8.9.10.11-7.8.9.10\)\(4C=8.9.10.11\)

\(C=\dfrac{8.9.10.11}{4}=1980.\)

c, https://hoc24.vn/hoi-dap/question/384591.html

Câu này bạn vào đây mình đã giải câu tương tự nhé.

\(1)A=\dfrac{5}{1.2}+\dfrac{5}{2.3}+...+\dfrac{5}{99.100}\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\left(1-\dfrac{1}{100}\right)\)

\(\Leftrightarrow A=5\cdot\dfrac{99}{100}\)

\(\Leftrightarrow A=\dfrac{99}{20}\)

3.đặt biểu thức trên là A

ta có:

2A=2.(\(\dfrac{1}{3.5}\)+\(\dfrac{1}{5.7}\)+\(\dfrac{1}{7.9}\)+...+\(\dfrac{1}{2015.2017}\))

=>2A=\(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)+\(\dfrac{2}{7.9}\)+....+\(\dfrac{2}{2015.2017}\)

=>2A=\(\dfrac{1}{3}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{7}\)+\(\dfrac{1}{7}\)-\(\dfrac{1}{9}\)+....\(\dfrac{1}{2015}\)-\(\dfrac{1}{2017}\)

=>2A=\(\dfrac{1}{3}\)-\(\dfrac{1}{2017}\)=\(\dfrac{2014}{6051}\)

=>A=\(\dfrac{2014}{6051}\):2=\(\dfrac{1007}{6051}\)

`# \text {Ryo}`

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}\\ =\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}\\ =\dfrac{1}{2}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\left(\dfrac{1}{11}-\dfrac{1}{11}\right)-\dfrac{1}{14}\\ =\dfrac{1}{2}-\dfrac{1}{14}\\ =\dfrac{7}{14}-\dfrac{1}{14}\\ =\dfrac{6}{14}\\ =\dfrac{3}{7}\)

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