Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{8.9.10}\right).x=\frac{23}{45}\)
\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{8.9}-\frac{1}{9.10}\right)\right]x=\frac{23}{45}\)
\(\Leftrightarrow\left[\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{9.10}\right)\right].x=\frac{23}{45}\)
\(\Leftrightarrow\left(\frac{1}{2}.\frac{44}{90}\right).x=\frac{23}{45}\)
\(\Leftrightarrow\frac{11}{45}.x=\frac{23}{45}\Rightarrow x=\frac{23}{45}:\frac{11}{45}=\frac{23}{11}\)
nhung sao banj khong phan h ra ro rang,chang nhe den do khong phan h duoc sao
a . Ta có : \(n+10⋮n+1\)
\(n+1+9⋮n+1\)
mà\(n+1⋮n+1\)
\(\Rightarrow9⋮n+1\)
\(\Rightarrow n+1\inƯ\left(9\right)=\left\{1;3;9\right\}\)
Ta có bảng sau :
| n +1 | 1 | 3 | 9 |
| n | 0 | 2 | 8 |
để n+10 chia hết n+1 thì
9chia hết cho n+1
=>n+1 \(\inƯ\left(9\right)=\left\{1;3;9\right\}\)
ta có bảng sau
| n+1 | 1 | 3 | 9 |
| n | 2 | 4 | 10 |
| tm | tm | tm |
vậy...
\(\Leftrightarrow x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}\right)=\frac{1}{100}+\frac{1}{99}-\frac{1}{100}\)
\(\Leftrightarrow x-\frac{98}{99}=\frac{1}{99}\Leftrightarrow x=1\)
\(A=355+\frac{354}{2}+\frac{353}{3}+...+\frac{2}{354}+\frac{1}{355}\)
\(A=1+\left(\frac{354}{2}+1\right)+...+\left(\frac{2}{354}+1\right)+\left(\frac{1}{355}+1\right)\)
\(A=1+\frac{356}{2}+...+\frac{356}{354}+\frac{356}{355}\)
\(A=\frac{356}{356}+\frac{356}{2}+...+\frac{356}{354}+\frac{356}{355}\)
\(A=356.\left(\frac{1}{2}+...+\frac{1}{354}+\frac{1}{355}+\frac{1}{356}\right)\)
Sorry , mk biết làm đến bước đấy thôi
Ta có : \(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
Đặt \(B=\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\)
Nên \(A< B\)
\(\Rightarrow A.B=\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\right)\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)
\(\Rightarrow A.B=\frac{1}{201}\)
Vì \(A< B\)
\(\Rightarrow A^2< A.B=\frac{1}{201}\)
\(\Rightarrow A^2< \frac{1}{201}\)
\(\RightarrowĐPCM\)
\(A=\frac{5}{2.1}+\frac{4}{1.11}+\frac{3}{11.2}+\frac{1}{2.15}+\frac{13}{15.4}\)
\(=\frac{1}{7}\left(\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\right)\)
\(=\frac{1}{7}\left(\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{28}\right)\)
\(=\frac{1}{2}.\frac{13}{28}\)
\(=\frac{13}{56}\)
Bạn qua đây mà hỏi nhé: Hỏi đáp môn Sinh học | Học trực tuyến
A = 1 . 2 . 3 + 2 . 3 . 4 + ......... + 38 .39 . 40
\(\Rightarrow4A=1.2.3.4+2.3.4.4+......+38.39.40.4\)
\(\Rightarrow4A=1.2.3.\left(4-0\right)+2.3.4.\left(5-1\right)+.......+38.39.40.\left(41-37\right)\)
\(\Rightarrow4A=1.2.3.4+1.2.3.0+2.3.4.5-1.2.3.4+....+38.39.40.41-38.39.40.37\)
\(\Rightarrow4A=38.39.40.41\)
\(\Rightarrow A=\frac{38.39.40.41}{4}\)
\(\Rightarrow A=38.39.10.41\)
\(\Rightarrow A=607620\)
Vậy \(A=607620\)
mơn bạn