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\(M=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{100.101.102}\right)\)
\(M=\frac{1}{2}.\left(1-\frac{1}{102}\right)\)
\(M=\frac{101}{204}< 1\left(đpcm\right)\)
Ta có: M=11.2.3 +12.3.4 +13.4.5 +...+1100.101.102
M=2.(11.2.3 +12.3.4 +13.4.5 +...+1100.101.102 ).12
M=(21.2.3 +22.3.4 +23.4.5 +...+2100.101.102 ).12
M=(11.2 -12.3 +12.3 -13.4 +13.4 -14.5 +...+1100.101 −1101.102 ).12
M=( 11.2 −1101.102 ).12
Mà
M=1/1x2x3 =1/2x3x4 +1/3x4x5 +..........+1/100x101x102
M=3-1/1x2x3 +4-2/2x3x4+5-3/3x4x5 + ......... +102-100/100x101x102
M=3/1x2x3 -1/1x2x3 +4/2x3x4 -2/2x3x4 +........... + 102/100x101x102 -100/100x101x102
M=1/1x2 -1/2x3 +1/2x3 -1/3x4 +......... + 1/100x101 -1/101x102
M=1/1x2 -1/101x102
M=2575/5151 < 1 suy ra M<1
Vậy M<1
\(M=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{100\cdot101\cdot102}\)
\(M=\frac{1}{2}\left[\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+...+\frac{2}{100\cdot101\cdot102}\right]\)
\(M=\frac{1}{2}\left[\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{100\cdot101}-\frac{1}{101\cdot102}\right]\)
\(M=\frac{1}{2}\left[\frac{1}{2}-\frac{1}{101\cdot102}\right]=\frac{1}{2}\cdot\frac{2575}{5151}=\frac{2575}{10302}< \frac{10302}{10302}=1\)
Vậy : M < 1
\(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{100.101.102}\)
\(M=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{100.101.102}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(M=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{102102}\right)\)
\(M=\frac{1}{2}.\left(\frac{51051}{102102}-\frac{1}{102102}\right)\)
\(M=\frac{1}{2}.\frac{25525}{51051}\)
\(M=\frac{25525}{102102}< 1\)
Vậy M<1
Ta thấy : \(\frac{1}{1\cdot2\cdot3}=\frac{1}{1\cdot3}-\frac{1}{2\cdot3};\frac{1}{2\cdot3\cdot4}=\frac{1}{2\cdot4}-\frac{1}{3\cdot4};...;\frac{1}{100\cdot101\cdot102}=\frac{1}{100\cdot102}-\frac{1}{101\cdot102}\)
\(\Rightarrow M=\frac{1}{1\cdot3}-\frac{1}{2\cdot3}+\frac{1}{2\cdot4}-\frac{1}{3\cdot4}+...+\frac{1}{100\cdot102}-\frac{1}{101\cdot102}\)
\(M=\left(\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+...+\frac{1}{100\cdot102}\right)-\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{101\cdot102}\right)\)
Đặt : \(A=\frac{1}{1\cdot3}+\frac{1}{2\cdot4}+...+\frac{1}{100\cdot102};B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{101\cdot102}\)
\(\Rightarrow2\cdot A=\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+...+\frac{2}{100\cdot102}\)
\(2\cdot A=\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{102}\)
\(2\cdot A=1+\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+...+\left(\frac{1}{100}-\frac{1}{100}\right)-\frac{1}{101}-\frac{1}{102}\)
\(2\cdot A=1+\frac{1}{2}-\frac{1}{101}-\frac{1}{102}\)
\(\Rightarrow A=\frac{1}{2}+\frac{1}{4}-\frac{1}{202}-\frac{1}{204};\)
\(B=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{101\cdot102}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{101}-\frac{1}{102}\)
\(B=\frac{1}{2}-\frac{1}{102}\)
Thay A, B vào M, ta có :
\(M=\left(\frac{1}{2}-\frac{1}{102}\right)-\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{202}-\frac{1}{204}\right)\)
\(M=\frac{1}{2}-\frac{1}{102}-\frac{1}{2}-\frac{1}{4}+\frac{1}{202}+\frac{1}{204}\)
\(M=\frac{1}{2}-\frac{1}{2}+\frac{1}{204}+\frac{1}{102}-\frac{1}{4}+\frac{1}{202}\)
\(M=\frac{1}{68}-\frac{1}{4}+\frac{1}{202}\)
\(M=-\frac{4}{17}+\frac{1}{202}hayM=\frac{1}{202}-\frac{4}{17}\)
Vì \(\frac{1}{202}< 1\Rightarrow\frac{1}{202}-\frac{1}{4}< 1\Rightarrow M< 1\left(đpcm\right)\)
Vậy: M < 1
Vậy là ai chậm nhất 9 tích à
Ủa vậy?Ai là đúng nhất vậy?
Tính tổng sau :
Z = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{48.49.50}\)
\(Z=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{48.49.50}\)
\(=\frac{1}{2}\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{48.49.50}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{48.49}-\frac{1}{49.50}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{49.50}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{2450}\right)=\frac{1}{2}\left(\frac{2450}{2450}-\frac{1}{2450}\right)\)
\(=\frac{1}{2}.\frac{2449}{2450}=\frac{2449}{4900}\)
Z = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/98.99.100
Áp dụng phương pháp khử liên tiếp: viết mỗi số hạng thành hiệu của hai số sao cho số trừ ở nhóm trước bằng số bị trừ ở nhóm sau.
Ta xét:
1/1.2 - 1/2.3 = 2/1.2.3; 1/2.3 - 1/3.4 = 2/2.3.4;...; 1/98.99 - 1/99.100 = 2/98.99.100
tổng quát: 1/n(n+1) - 1/(n+1)(n+2) = 2/n(n+1)(n+2). Do đó:
2Z = 2/1.2.3 + 2/2.3.4 + 2/3.4.5 +...+ 2/98.99.100
= (1/1.2 - 1/2.3) + (1/2.3 - 1/3.4) +...+ (1/98.99 - 1/99.100)
= 1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + ... + 1/98.99 - 1/99.100
= 1/1.2 - 1/99.100
= 1/2 - 1/9900
= 4950/9900 - 1/9900
= 4949/9900.
Vậy Z = \(\frac{4949}{9900}\)
Rút gọn biểu thức sau:
S = \(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+\(\frac{1}{3.4.5}\)+ ... + \(\frac{1}{998.999.1000}\)
Giải :
\(\text{S}=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{998\cdot999\cdot1000}\)
\(\text{S}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{998}-\frac{1}{999}+\frac{1}{999}-\frac{1}{1000}\)
\(\text{S}=1-\frac{1}{1000}=\frac{999}{1000}\)
\(S=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{998.999.1000}\)
\(=\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{1000-998}{998.999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{998.999}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{999.1000}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{999000}\right)\)
\(=\frac{1}{2}.\frac{499499}{999000}\)
\(=\frac{499499}{1998000}\)
Study well ! >_<
Tính : \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+....+\frac{1}{98.99.100}\)
Kết bạn vs mk nha !!!
Tính giá trị biểu thức sau:
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+.....+\frac{1}{18.19.20}\)
Giúp mình với!!!!Đang cần gấp
Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)
\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)
\(\Rightarrow B=\frac{189}{760}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)
\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)
S = \(\frac{9^{2018}+1}{9^{2017}+1}\)và M = \(\frac{9^{2017}+1}{9^{2016}+1}\)
Hãy so sánh nha !
Các bạn giúp mình nha ai nhanh mình tick cho !
Ta có \(\frac{1}{9S}=\frac{9^{2017}+\frac{1}{9}}{9^{2017}+1}\)= \(\frac{9^{2017}+1-\frac{8}{9}}{9^{2017}+1}=1-\frac{\frac{8}{9}}{9^{2017}+1}\)
\(\frac{1}{9M}=\frac{9^{2016}+\frac{1}{9}}{9^{2016}+1}\)= \(\frac{9^{2016}+1-\frac{8}{9}}{9^{2016}+1}=1-\frac{\frac{8}{9}}{9^{2016}+1}\)
Vì \(9^{2016}+1< 9^{2017}+1\)=> \(\frac{\frac{8}{9}}{9^{2016}+1}>\frac{\frac{8}{9}}{9^{2017}+1}\)
=> \(1-\frac{\frac{8}{9}}{9^{2016}+1}< 1-\frac{\frac{8}{9}}{9^{2017}+1}\)=> \(\frac{1}{9}S< \frac{1}{9}M\Rightarrow S< M\)
Các Bạn ơi mk với mk cần gấp:
Bài 1: Tính giá trị biểu thức sau:
A= -\(\frac{11}{23}\). \(\frac{10}{-13}\) +\(\frac{-11}{13}\) .\(\frac{3}{-23}\) -\(\frac{2}{23}\) C=\(\frac{5^2}{1.6}\)+\(\frac{5^2}{6.11}\) +\(\frac{5^2}{11.16}\) +. . . +\(\frac{5^2}{31.36}\)
B=\(\frac{1}{1.2.3}\) +\(\frac{1}{2.3.4}\) +\(\frac{1}{3.4.5}\) + . . . .+\(\frac{1}{18.19.20}\) D=\(\frac{1.2.3+2.4.6+4.8.12+7.14.21}{1.3.5+2.6.10+4.12.20+7.21.35}\) +\(\frac{3}{5}\)
Tính gt biểu thức :
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+.......+\frac{1}{18.19.20}\)
nhớ trình bày cả lời giải nha !
Ai nhanh milk tick!
trình bày cụ thể hơn nha bạn
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
= \(\frac{1}{1}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{20}\)
=\(\frac{1}{1}-\frac{1}{20}=\frac{19}{20}\)
Tính giá trị A,B,C bt:
a,A=\(\frac{2.3.4+4.6.8+8.12.16+8.32.48}{3.4.5+6.8.10+12.16.20+24.32.40}\)
b,B=\(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
c,C=1+1/2+1/2.3+1/3.4+...+1/2013.2014+1/2014.2015
Các bn ới giúp mik với
Ai đúng mik cho 3 tik mỗi ý tương đương 1 tick
So sánh A và B biết:
A= \(\frac{1+5+5^2+......+5^9}{1+5+5^2+......+5^8}\)
B= \(\frac{1+3+3^2+.....+3^9}{1+3+3^2+.....+3^8}\)
Ai nhanh mk tick!
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