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a/ \(A=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+........+\frac{99}{100!}\)
\(\Leftrightarrow A=\frac{2-1}{2!}+\frac{3-1}{3!}+\frac{4-1}{4!}+......+\frac{100-1}{100!}\)
\(\Leftrightarrow A=\frac{2}{2!}-\frac{1}{2!}+\frac{3}{3!}-\frac{1}{3!}+.....+\frac{100}{100!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+....+\frac{1}{99!}-\frac{1}{100!}\)
\(\Leftrightarrow A=1-\frac{1}{100!}\)
b/ \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+.....+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{9900}\)
Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)
\(\Rightarrow B=\frac{189}{760}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}.\frac{22}{45}\)
\(=\frac{11}{45}\)
Mk ko ghi lại đề nha :3
\(\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}-\frac{1}{10}\)
=\(=\frac{1}{1}-\frac{1}{10}=\frac{10-1}{10}=\frac{9}{10}\)
\(\left(\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-.........+\frac{1}{8}-\frac{1}{9}-\frac{1}{10}\right)\).x = \(\frac{22}{45}\)
\(\left(1-\frac{1}{10}\right).x=\frac{22}{45}\)
\(\frac{9}{10}.x=\frac{22}{45}\)
\(x=\frac{22}{45}:\frac{9}{10}\)
\(x=\frac{44}{81}\)
Theo đầu bài ta có:
\(\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\cdot x=\frac{23}{45}\)
\(\Rightarrow\frac{\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}}{2}\cdot x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{1\cdot2}-\frac{1}{9\cdot10}\right)\cdot x=\frac{46}{45}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)\cdot x=\frac{46}{45}\)
\(\Rightarrow\frac{22}{45}\cdot x=\frac{46}{45}\)
\(\Rightarrow x=\frac{23}{11}\)
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\)
\(A=\frac{1}{2}.\frac{22}{45}=\frac{11}{45}\)
\(\Rightarrow\frac{11}{45}.x=\frac{23}{45}\)
\(\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
Ủng hộ mk nha !!! ^_^
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+..........+\frac{1}{8.9}-\frac{1}{9.10}\)
\(=\frac{1}{1.2}-\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{90}\)
\(=\frac{45}{90}-\frac{1}{90}\)
\(=\frac{44}{90}\)
\(=\frac{22}{45}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)
\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)
\(\Leftrightarrow\left(\frac{1}{2}.\frac{44}{90}\right)x=\frac{23}{45}\)
\(\Leftrightarrow\frac{11}{45}x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{8.9.10}\right)x=\frac{23}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)x=\frac{48}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{48}{45}\)
\(\Rightarrow\frac{22}{45}x=\frac{48}{45}\)
\(\Rightarrow x=\frac{24}{11}\)
Vậy...
Ta có : 1/1 -1/2 -1/3 +1/2-1/3-1/4+........+1/8-1/9-1/10
Ta gạch các phân số ở giữa còn lại 1/1-1/10=9/10
Giải :
\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\)
\(=\frac{1}{2}-\frac{1}{9\cdot10}=\frac{1}{2}-\frac{1}{90}\)
\(=\frac{45}{90}-\frac{1}{90}=\frac{44}{90}=\frac{22}{45}\)
\(@Cothanhkhe-hoqchac\)
sai rồi liu liu
\(\frac{1}{1}-\frac{1}{10}\)làm phép tính này là ra kết quả
con gà cố thanh khê
\(=1-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}-\frac{2}{10}\)
\(=\)\(1-\frac{1}{10}\)
\(=\frac{9}{10}\)
\(A=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+....+\frac{1}{8\cdot9\cdot10}\)
\(\Rightarrow2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+....+\frac{2}{8\cdot9\cdot10}\)
\(\Rightarrow2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{90}\)
\(\Rightarrow2A=\frac{44}{90}\)
\(\Rightarrow A=\frac{44}{180}\)
CTv còn làm sai
thanh khê bị ngu rồi
gọi biểu thức trên là A
=> A =1/2.(1/1.2-1/2.3+1/2.3-1/3.4+...+1/8.9-1/9.10)
=> A= 1/2. (1/1.2-1/9.10)
=> A = 1/2. (1/2-1/90)
=> A = 1/2. ( 45/90-1/90)
=> A = 1/2 .44/90
=> A = 88/90
=> A = 44/45
=> A =
Nguyen Duy Anh:Thưa cao nhân.em làm sai chỗ nào vậy ạ ?
sai bét
Em học từ lâu rôi anh ạ anh bị ngu à
Nguyen 3D sai chỗ nào?
ko biết
Nguyen Duy Anh sai chỗ nào
Làm chi tiết nha
Câu hỏi của Kim Sura xXx pÉ heO:Hi vọng bạn xem qua bài này thì bớt sure conmej bạn ạ.
Nhân dân có câu uốn lưỡi 7 lần trc khi nói mà bạn đã sure 10 lần trc khi comment chưa ạ.nếu có ý kiến cho rằng mik sai thì ib riêng nhé!tớ ko mún làm nhiễu diễn đàn!
Sukvatsua'sss
sai ở chỗ là \(\frac{1}{1}-\frac{1}{10}\)
Câu hỏi của Kim Sura xXx pÉ heO
\(\frac{1}{1}-\frac{1}{10}\)sai chỗ nào?
Sửa :
\(2A=2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{8\cdot9\cdot10}\right)\)
\(2A=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{8\cdot9\cdot10}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{8\cdot9}-\frac{1}{9\cdot10}\)
\(2A=\frac{1}{1\cdot2}-\frac{1}{9\cdot10}\)
\(2A=\frac{1}{2}-\frac{1}{90}\)
\(2A=\frac{45}{90}-\frac{1}{90}=\frac{44}{90}\)
\(A=\frac{44}{90}\div2=\frac{44}{90}\cdot\frac{1}{2}=\frac{44}{180}\)
\(@Cothanhkhe\)
Đây nhé,t chứng minh công thức tổng quát luôn khỏi cãi.
Ta sẽ c/m: \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) (n thuộc N*)
Thật vậy,ta biến đổi vế phải:\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Từ đây suy ra \(VP=\frac{1}{2}.\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)\left(n+2\right)}=VT^{\left(đpcm\right)}\)
Áp dụng vào bài toán ta có:
\(VT=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{90}\right)=\frac{44}{180}=\frac{11}{45}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{8.9.10}\)
\(=\frac{1}{1}-\frac{1}{2}-\frac{1}{3}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+...+\frac{1}{8}-\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{1}-\frac{1}{10}\)
\(=\frac{10}{10}-\frac{1}{10}\)
\(=\frac{9}{10}\)