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(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=\(\dfrac{3.5+99.101.103.105}{8}=13517400\)
\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=4.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=4.\frac{3200}{9603}=\frac{12800}{9603}\)
\(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{95.97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{95.97}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{97.99}\right)\)
\(A=\frac{1}{4}.\frac{3200}{9603}\)
\(A=\frac{800}{9603}\)
Bài trc mik làm lộn :)))
~ Hok tốt ~
A = 1/4.( 4/1.3.5 + 4/3.5.7+ ....+ 4/95.97.99)
= 1/4 .( 1/ 1.3 - 1/3.5 + 1/3.5 - 1/5.7 + .......+ 1/95.97 - 1/97.99)
= 1/4( 1/1.3 - 1/97.99)
= 1/4 . 9499/29397
\(\Rightarrow4x-\left(\frac{4}{5.7.9}+\frac{4}{7.9.11}+...+\frac{4}{99.101.103}\right)=\frac{2}{83224}=\frac{1}{41612}\)
\(4x-\left(\frac{9-5}{5.7.9}+\frac{11-7}{7.9.11}+...+\frac{103-99}{99.101.103}\right)=\frac{1}{41612}\)
\(4x-\left(\frac{1}{5.7}-\frac{1}{7.9}+\frac{1}{7.9}-\frac{1}{9.11}+...+\frac{1}{99.101}-\frac{1}{101.103}\right)=\frac{1}{41612}\)
\(4x-\left(\frac{1}{5.7}-\frac{1}{101.103}\right)=\frac{1}{41612}\)
Từ đó tìm ra x
a) \(A=\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{31.33.35}\)
=>\(2A=\frac{2}{3.5.7}+\frac{2}{5.7.9}+...+\frac{2}{31.33.35}\)
\(=\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{31.33}-\frac{1}{33.35}\)
\(=\frac{1}{15}-\frac{1}{1155}=\frac{77}{1155}-\frac{1}{1155}=\frac{76}{1155}\)
=> \(A=\frac{76}{1155}:2=\frac{76}{1155}.\frac{1}{2}=\frac{38}{1155}\)
b)
\(B=\frac{1}{1.4.7}+\frac{1}{7.10.13}+...+\frac{1}{54.57.60}\)
sai đề thì phải
ko sai thì bạn làm đi Tra