TN
1
K
Khách
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P
2
HT
24 tháng 7 2015
A = \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{2499}{2500}\)
A = \(\frac{1.3.2.4.3.5.......49.51}{2.2.3.3.4.4........50.50}\)
A = \(\frac{\left(1.2.3.......49\right).\left(3.4.5.....51\right)}{\left(2.3.4.....50\right)\left(2.3.4.....50\right)}\)
A = \(\frac{51}{50.2}\)
A = \(\frac{51}{100}\)
24 tháng 7 2015
A = \(\frac{1.3}{2.2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\cdot\cdot\cdot\cdot\frac{49\cdot51}{50\cdot50}=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot49\cdot51}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot\cdot\cdot\cdot50\cdot51}=\frac{1\cdot51}{2\cdot50}=\frac{51}{100}\)
NT
1
14 tháng 3 2018
a=8/9+15/16+24/25+....+2499/2500
a=(1-1/9)+(1-1/16)+(1-1/25)+....+(1-1/2500)
a=1-1/9+1-1/16+1-1/25+....+1-1/2500
a=(1+1+...+1)-(1/9+1/16+1/25+....+1/2500)
NK
0
HM
0
DG
7
VA
14 tháng 4 2017
\(A=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(=\frac{1.3}{2^2}+\frac{2.4}{3^2}+\frac{3.5}{4^2}+...+\frac{49.51}{50^2}\)
\(=\frac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)
\(=\frac{\left(1.2.3...49\right)\left(3.4.5...51\right)}{2^2.3^2.4^2...50^2}\)
\(=\frac{1.2.50.51}{2^2.50^2}=\frac{51}{100}\)
T
19 tháng 6 2019
#)Giải :
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{2499}{2500}\)
\(A=\frac{1.3}{2.2}\times\frac{2.4}{3.3}\times\frac{3.5}{4.4}\times\frac{4.6}{5.5}\times...\times\frac{49.51}{50.50}\)
\(A=\frac{1\times3\times2\times4\times3\times5\times...\times49\times51}{2\times2\times3\times3\times4\times4\times...\times50\times50}\)
\(A=\frac{1\times51}{2\times50}\)
\(A=\frac{51}{100}\)
19 tháng 6 2019
\(A=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times\frac{24}{25}\times...\times\frac{2499}{2500}\)
\(=\frac{1\times3}{2\times2}\times\frac{2\times4}{3\times3}\times\frac{3\times5}{4\times4}\times\frac{6\times4}{5\times5}\times...\times\frac{49.51}{50\times50}\)
\(=\frac{1}{2}\times\frac{51}{50}\)
\(=\frac{51}{100}\)
DL
1 tháng 4 2018
\(A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{2499}{2500}\)
\(A=\frac{3}{2^2}\cdot\frac{8}{3^2}\cdot\frac{15}{4^2}\cdot....\cdot\frac{2499}{50^2}\)
\(A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot\frac{3\cdot5}{4\cdot4}\cdot...\cdot\frac{49\cdot51}{50\cdot50}\)
\(A=\frac{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot49\cdot51}{2\cdot2\cdot3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot50\cdot50}\)
\(A=\frac{1\cdot51}{2\cdot50}=\frac{51}{100}\)
WH
1 tháng 4 2018
\(A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot\cdot\cdot\cdot\cdot\frac{2499}{2500}\)
\(\Rightarrow A=\frac{1.3}{2\cdot2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{49.51}{50.50}\)
\(\Rightarrow A=\frac{1.3.2.4.3.5.....49.51}{2.2.3.3.4.4.....50.50}\)
\(\Rightarrow A=\frac{\left(1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot49\right)\cdot\left(2\cdot3\cdot4\cdot\cdot\cdot\cdot\cdot51\right)}{\left(3\cdot4\cdot\cdot\cdot\cdot\cdot50\right)\cdot\left(2\cdot3\cdot\cdot\cdot\cdot\cdot\cdot50\right)}\)
\(\Rightarrow A=\frac{1.51}{2\cdot50}\)
\(\Rightarrow A=\frac{51}{100}\)
KD
1
6 tháng 2 2017
\(B=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}\)
\(\Rightarrow B=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{2500}\right)\)
\(\Rightarrow B=\left(1-\frac{1}{2^2}\right)+\left(1-\frac{1}{3^2}\right)+\left(1-\frac{1}{4^2}\right)+...+\left(1-\frac{1}{50^2}\right)\)
\(\Rightarrow B=\left(1+1+1+...+1\right)-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\) (có 49 số 1)
\(\Rightarrow B=49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1-\frac{1}{50}\)<1
\(\Rightarrow-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>-1\)
\(\Rightarrow49-\left(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}\right)>49-1\)
\(\Rightarrow B>48\)
NT
0
\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{49.51}{50^2}\)
\(A=\frac{1.2.3...49}{2.3.4...50}.\frac{3.4.5...51}{2.3.4...50}\)
\(A=\frac{1}{50}.\frac{51}{2}\)
\(A=\frac{51}{100}\)