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\(\frac{3}{1.3}+\frac{3}{3.5}+.....+\frac{3}{49.51}\)
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NT
2
11 tháng 5 2017
\(A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\)
\(A=1-\frac{1}{51}\)
\(A=\frac{50}{51}\)
Q
11 tháng 5 2017
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(2A=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(2A=3\left(1-\frac{1}{51}\right)\)
\(2A=3.\frac{50}{51}\)
\(2A=\frac{50}{17}\Rightarrow A=\frac{25}{17}\)'
LN
1
18 tháng 6 2020
ta có A=3/1*3+3/3*5+3/5*7+...+3/49*51
=> A=3*1/2*(2/1*3+2/3*5+..+2/49*51)
=> A=3/2*(1-1/3+1/3-1/5+..+1/49-1/51)
=> A=3/2*(1-1/51)
=> A= 3/2* 50/51
=> A= 25/17
NV
2
25 tháng 4 2018
Ta có :
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
Vậy \(A=\frac{25}{17}\)
Chúc bạn học tốt ~
25 tháng 4 2018
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
7 tháng 8 2016
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
HT
7 tháng 8 2016
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=3.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{49}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\frac{49}{50}\)
\(A=\frac{147}{100}\)
TG
4
23 tháng 7 2016
\(\text{= 2/1 . 2/3 . 3/2 . 3/4 . 4/3 . 4/5 ....... 50/49.50/51 }\)
Dùng phương pháp khử liên tiếp ta có
\(=\frac{2}{1}-\frac{50}{51}=\frac{52}{51}\)
2 tháng 4 2016
A=3/1.3+3/3.5+3/5.7+............+3/49.51
A=3/1-3/3=3/3-3/5+3/5-3/7+...............+3/49-3/51
A=1-1/3+1/3-1/5+1/5-1/7+.....................+1/39-1/51
A=1-1/51
A=50/51
HV
2 tháng 4 2016
A\(=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...\frac{1}{49.51}\right) \)
\(=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...\frac{2}{49.51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
=\(\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(=\frac{3}{2}.\frac{50}{51}\)
\(=\frac{25}{17}\)
24 tháng 3 2019
\(\Leftrightarrow N=\frac{\left(2.3.4....50\right)\left(2.3.4...........50\right)}{\left(1.2.3.........49\right)\left(3.4.5...........51\right)}=\frac{50.2}{51}=\frac{100}{51}\)
25 tháng 3 2019
\(\frac{2^2}{1.3}+\frac{3^2}{2.4}+\frac{4^2}{3.5}+....+\frac{50^2}{49.51}\)
\(=\frac{2^2-1}{1.3}+\frac{3^2-1}{2.4}+....+\frac{50^2-1}{49.51}+\frac{1}{1.3}+\frac{1}{2.4}+....+\frac{1}{49.51}\)
\(=\frac{1}{2}.\left(1+1+...+1\right)+\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{49}-\frac{1}{51}\)
Tự làm tiếp :))
25 tháng 3 2019
tớ nhầm đoạn này tí :((
\(=\left(1+1+....+1\right)+\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)(49 chữ số 1)
\(=49+\frac{1}{2}.\left[\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{49}\right)-\left(\frac{1}{3}+\frac{1}{4}+...+\frac{1}{51}\right)\right]\)
\(=49+\left(\frac{3}{2}-\frac{1}{50}-\frac{1}{51}\right):2\)Tự tính
\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}=\frac{3}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\left(1-\frac{1}{51}\right)=\frac{3}{2}.\frac{50}{51}=\frac{25}{17}\)
\(\frac{3}{1.3}+\frac{3}{3.5}+...+\frac{3}{49.51}=\frac{3}{2}\cdot\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{49.51}\right)\)
\(=\frac{3}{2}\cdot\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(=\frac{3}{2}\cdot\left(1-\frac{1}{51}\right)=\frac{3}{2}\cdot\frac{50}{51}=\frac{3.50}{2.51}=\frac{1.25}{1.17}=\frac{25}{17}\)