c=94/285

mình cũng ko hiểu cách giải này cho lắm nên mình làm thế thôi

kq 47/190

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
=68(81.3.7+83.7.9+...+813.15.19)=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)=​8​​6​​(​1.3.7​​8​​+​3.7.9​​8​​+...+​13.15.19​​8​​)
=68(11.3−13.7+13.7−17.9+...+113.15−115.19)=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)=​8​​6​​(​1.3​​1​​−​3.7​​1​​+​3.7​​1​​−​7.9​​1​​+...+​13.15​​1​​−​15.19​​1​​)
=68⋅(13−1285)=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)=​8​​6​​⋅(​3​​1​​−​285​​1​​)
=68⋅94285=\frac{6}{8}\cdot\frac{94}{285}=​8​​6​​⋅​285​​94​​
=47190=\frac{47}{190}=​190​​47​​

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)=​8​​6​​(​1.3.7​​8​​+​3.7.9​​8​​+...+​13.15.19​​8​​)
=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)=​8​​6​​(​1.3​​1​​−​3.7​​1​​+​3.7​​1​​−​7.9​​1​​+...+​13.15​​1​​−​15.19​​1​​)
=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)=​8​​6​​⋅(​3​​1​​−​285​​1​​)
=\frac{6}{8}\cdot\frac{94}{285}=​8​​6​​⋅​285​​94​​
=\frac{47}{190}=​190​​47​​

B=6/1.3.5+6/3.7.9+6/7.9.13+6/9.13.15+6/13.15.19

 =2/5+2/63+2/273+2/585+2/1235

 =886/1995

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19

\(=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)\)

\(=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)\)

\(=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)\)

\(=\frac{6}{8}\cdot\frac{94}{285}\)

\(=\frac{47}{190}\)

6/1.3.7 + 6/3.7.9 + 6/7.9.13 + 6/9.13.15 + 6/13.15.19
=\frac{6}{8}\left(\frac{8}{1.3.7}+\frac{8}{3.7.9}+...+\frac{8}{13.15.19}\right)=​8​​6​​(​1.3.7​​8​​+​3.7.9​​8​​+...+​13.15.19​​8​​)
=\frac{6}{8}\left(\frac{1}{1.3}-\frac{1}{3.7}+\frac{1}{3.7}-\frac{1}{7.9}+...+\frac{1}{13.15}-\frac{1}{15.19}\right)=​8​​6​​(​1.3​​1​​−​3.7​​1​​+​3.7​​1​​−​7.9​​1​​+...+​13.15​​1​​−​15.19​​1​​)
=\frac{6}{8}\cdot\left(\frac{1}{3}-\frac{1}{285}\right)=​8​​6​​⋅(​3​​1​​−​285​​1​​)
=\frac{6}{8}\cdot\frac{94}{285}=​8​​6​​⋅​285​​94​​
=\frac{47}{190}=​190​​47​​

Giải:

Đặt:

\(A=\dfrac{6}{1.3.7}+\dfrac{6}{3.7.9}+\dfrac{6}{7.9.13}+\dfrac{6}{9.13.15}+\dfrac{6}{13.15.19}\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{8}{1.3.7}+\dfrac{8}{3.7.9}+\dfrac{8}{7.9.13}+\dfrac{8}{9.13.15}+\dfrac{8}{13.15.19}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{1.3}-\dfrac{1}{3.7}+\dfrac{1}{3.7}-\dfrac{1}{7.9}+\dfrac{1}{7.9}-\dfrac{1}{9.13}+\dfrac{1}{9.13}-\dfrac{1}{13.15}+\dfrac{1}{13.15}-\dfrac{1}{15.19}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{1.3}-\dfrac{1}{15.19}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}\left(\dfrac{1}{3}-\dfrac{1}{285}\right)\)

\(\Leftrightarrow A=\dfrac{6}{8}.\dfrac{94}{285}\)

\(\Leftrightarrow A=\dfrac{47}{190}\)

Vậy ...

 D = 4/1.3.5+4/3.5.7+4/5.7.9+4/7.9.11+4/9.11.13

D.\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{9.11}-\frac{1}{11.13}\)

D\(=\frac{1}{1.3}-\frac{1}{11.13}\)

D\(=\frac{1}{3}-\frac{1}{143}\Rightarrow D=\frac{140}{429}\)

(chỉnh đề)

A=\(-1+2-3-4-5+6-7-8-9+...-2021-2022+2023-2024\)

=\(\left(-1-2024\right)+\left(2+2023\right)+\left(-3-2022\right)+\left(-4-2021\right)+\left(-5-2020\right)+\left(6+2019\right)-\left(-7-2018\right)+\left(-8-2017\right)+\left(-9-2016\right)+...+\left(1010+1015\right)+\left(-1011-1014\right)+\left(-1012-1013\right)\)=\(-2025+2025-2025-2025-2025+2025-2025-2025-2025+...+2025-2025-2025\)=253.2025-1771.2025=-3 073 950.

B=\(1.3.5+3.5.7+5.7.9+7.9.11+...+99.101.103\)

8B=\(1.3.5.8+3.5.7.8+5.7.9.8+7.9.11.8+...+99.101.103.8\)

8B=\(1.3.5.\left[7-\left(-1\right)\right]+3.5.7.\left(9-1\right)+5.7.9.\left(11-3\right)+7.9.11.\left(13-5\right)+...+99.101.103.\left(105-97\right)\)8B=\(3.5+3.5.7+3.5.7.9-3.5.7+5.7.9.11-3.5.7.9+7.9.11.13-5.7.9.11+...+99.101.103.105-97.99.101.103\)

B=\(\dfrac{3.5+99.101.103.105}{8}=13517400\)

- Đặt C₁=\(1.3+3.5+...+197.199\).

6C₁=\(1.3.6+3.5.6+...+197.199.6\)

6C₁=\(1.3.\left[5-\left(-1\right)\right]+3.5.\left(7-1\right)+...+197.199.\left(201-195\right)\)

6C₁=\(3+3.5+3.5.7-3.5+...+197.199.201-195.197.199\)
C₁=\(\dfrac{3+197.199.201}{6}=1313301\)

- Đặt C₂=\(2.4+4.6+...+198.200\)

6C₂=\(2.4.6+4.6.6+...+198.200.6\)

6C₂=\(2.4.\left(6-0\right)+4.6.\left(8-2\right)+...+198.200.\left(202-196\right)\)

C₂=\(\dfrac{198.200.202}{6}=1333200\)

=>C₁+C₂=C=1313301+1333200=2646501