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\(M=\frac{5}{1.2.3}+\frac{5}{2.3.4}+\frac{5}{3.4.5}+...+\frac{5}{10.11.12}\)
\(=\frac{5}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+....+\frac{2}{10.11.12}\right)\)
\(=\frac{5}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(=\frac{5}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(=\frac{5}{2}.\frac{65}{132}=\frac{325}{264}\)
\(=\frac{5}{3}\left(\frac{1}{1\times2}-\frac{1}{2\times3}+...+\frac{1}{10\times11}-\frac{1}{11\times12}\right)\)
\(=\frac{5}{3}\times\left(\frac{1}{1\times2}-\frac{1}{11\times12}\right)\)
\(=\frac{5}{3}\times\left(1-\frac{1}{2}+\frac{1}{11}-\frac{1}{12}\right)\)
\(=\frac{5}{3}\times\frac{67}{132}\)
\(=\frac{335}{396}\)
\(2M=2\cdot\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+.....+\frac{1}{10\cdot11\cdot12}\right)\)
\(2M=\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+.....+\frac{2}{10\cdot11\cdot12}\)
\(2M=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+.....+\frac{1}{10\cdot11}-\frac{1}{11\cdot12}\)
\(2M=\frac{1}{1\cdot2}-\frac{1}{11\cdot12}\)
\(2M=\frac{1}{2}-\frac{1}{132}\)
\(2M=\frac{66}{132}-\frac{1}{132}\)
\(2M=\frac{65}{132}\)
\(M=\frac{65}{132}:2\)
\(M=\frac{65}{264}\)
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+.....+\frac{1}{10.11}-\frac{1}{11.12}\)
\(M=\frac{1}{2}-\frac{1}{11.12}=\frac{65}{132}\)
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{10.11.12}\)
\(M=\frac{1}{2}-\frac{1}{11.12}\)
\(M=\frac{65}{132}\)
Ngắn gọn , xúc tích !!! :))
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
đặt S=1.2.3+2.3.4+....+47.48.49
4S=1.2.3.(4-0)+2.3.4.(5-1)+...+47.48.49.(50-46)
4S=1.2.3.4-1.2.3+2.3.4.5-1.2.3.4+....+47.48.49.50-46.47.48.49
4S=47.48.49.50-1.2.3
S=(47.48.49.50-1.2.3):4
gọi biểu thức là A
ta có :
A=3/1.2.3 + 5/2.3.4 + 7/3.4.5 +....+ 2017/1008.1009.1010
A= (1.2/1.2.3 + 2.2/2.3.4 + 3.2/3.4.5 + ... + 1008.2/1008.1009.1010) + (1/1.2.3 + 1/2.3.4 + 1/3.4.5 +...+ 1/1008.1009.1010)
A=(2/2.3 + 2/3.4 + 2/4.5 +...+ 2/1009.1010 + 1/2.(1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5 + ... + 1/1008.1009 - 1/1009.1010
A=2(1/2-1/3+1/3-1/4+1/4-1/5+...+1/1009-1/1010)+1/2.(1/2-1/1009.1/1010)
A<2.1/2 + 1/2.1/2 = 1+1/4 = 5/4
OK nhớ tk cho mình nhé ( dấu này / là dấu phần nhé) chúc bạn học tốt
\(M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(=\frac{1}{2}-\frac{1}{11.12}\)
\(=\frac{65}{132}\)
Ta có nhận xét: 1/1.2 - 1/2.3 = 3-1/1.2.3 = 2/1.2.3
1/2.3 - 1/3.4 = 4-2/2.3.4 = 2/2.3.4
Suy ra: 1/1.2.3 = 1/2(1/1.2 - 1/2.3)
1/2.3.4 = 1/2(1/2.3 -1/3.4)
Do đó: M = 1/2(1/1.2-1/2.3 + 1/2.3 -1/3.4 + ... + 1/10.11 -1/11.12)
= 1/2(1/1.2 - 1/11.12) = 1/2(1/2-11/12 )
= 1/2.65/132 = 65/264
Phức tạp lắm
2P=2/1.2.3+2/2.3.4+2/3.4.5+2/10.11.12
2P=1/1.2-1/2.3+1/2.3-1/3.4+1/3.4-1/4.5+.....+1/10.11-1/11.12
2P=1/1.2-1/11.12
2P=1/2-1/132
2P=66/132-1/132
2P=65/132
P=65/264
\(P=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{10.11.12}\)
\(P=\dfrac{1}{2}-\dfrac{1}{11.12}\)
\(P=\dfrac{65}{132}\)
Ta có :
\(M=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{10.11.12}\)
\(M=5.\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{10.11.12}\right)\)
\(M=5.\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\right)\)
\(M=\frac{5}{2}.\left(\frac{1}{1.2}-\frac{1}{11.12}\right)\)
\(M=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{132}\right)\)
\(M=\frac{5}{2}.\left(\frac{66}{132}-\frac{1}{132}\right)\)
\(M=\frac{5}{2}.\frac{65}{132}\)
\(M=\frac{325}{264}\)
Tham khảo nha !!! Chúc học tốt !!!
Công thức :
\(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}\right)=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{6}\right)=\frac{1}{2}.\frac{1}{3}=\frac{1}{1.2.3}\)