K
Khách
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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
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(A=\frac{3}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(A=\frac{3}{2}\left(1-\frac{1}{51}\right)\)
\(A=\frac{3}{2}.\frac{50}{51}\)
\(A=\frac{25}{17}\)
\(B=\frac{21}{4}\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(B=\frac{21}{4}\left(\frac{33}{3.4}+\frac{33}{4.5}+\frac{33}{5.6}+\frac{33}{6.7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(B=\frac{21}{4}.33.\frac{4}{21}\)
\(B=\left(\frac{21}{4}.\frac{4}{21}\right).33\)
\(B=33\)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(C=\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(C=\frac{1}{2}\left(1-\frac{1}{99}\right)\)
\(C=\frac{1}{2}.\frac{98}{99}\)
\(C=\frac{49}{99}\)
DN
0
B
3
3 tháng 8 2018
\(P=\frac{3}{1.5}+\frac{3}{5.9}+\frac{3}{9.13}+...+\frac{3}{197.201}\)
\(P=\frac{3}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{197.201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}+\frac{1}{13}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\left(\frac{201}{201}-\frac{1}{201}\right)\)
\(P=\frac{3}{4}.\frac{200}{201}\)
\(P=\frac{50}{67}\)
Vậy \(P=\frac{50}{67}\)
3 tháng 8 2018
\(P=\frac{3}{1\cdot5}+\frac{3}{5\cdot9}+...+\frac{3}{197\cdot201}\)
\(=3\cdot\left(\frac{1}{1\cdot5}+\frac{1}{5\cdot9}+...+\frac{1}{197\cdot201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{4}{1\cdot5}+\frac{4}{5\cdot9}+...+\frac{4}{197\cdot201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{1}{1}-\frac{1}{201}\right)\)
\(=\frac{3}{4}\cdot\left(\frac{201-1}{201}\right)\)
\(=\frac{3}{4}\cdot\frac{200}{201}\)
\(\Rightarrow B=\frac{50}{67}\)
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
\(A=\frac{3}{1.3}+\frac{3}{3.5}+\frac{3}{5.7}+...+\frac{3}{49.51}\)
\(2A=3\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=3\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)\)
\(2A=3\left(1-\frac{1}{51}\right)\)
\(2A=3.\frac{50}{51}\)
\(2A=\frac{50}{17}\Rightarrow A=\frac{25}{17}\)'
C
7
8 tháng 4 2017
1/2+2/3+3/4+4/5+5/6+6/7+7/8+8/9+9/10x9/10
=9/10x(1/2+2/3)+(3/4+4/5)+(5/6+6/7)+(7/8+8/9)
=9/10x(1/3+3/5+5/7+7/9)
9/10x(1/3+3/5)+(5/7+7/9)
=9/10x1/5+5/9
9/50+5/9
=10
8 tháng 10 2017
Bn Long làm đúng rồi bn 
nguyễn kim arica cứ làm theo cách đó là được .
Bn nào thấy đúng thì ủng hộ nha .
NC
1
HP
10 tháng 6 2016
\(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^8}\)
\(=>3A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^7}\)
\(=>3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(=>2A=1-\frac{1}{3^8}=>A=\left(1-\frac{1}{3^8}\right):2\)
HT
18 tháng 8 2015
C = \(\frac{3}{2.3.4}+\frac{3}{3.4.5}+...+\frac{3}{98.99.100}\)
C = \(3.\frac{1}{2}.\left(\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{100-98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{4}{2.3.4}-\frac{2}{2.3.4}+\frac{5}{3.4.5}-\frac{3}{3.4.5}+...+\frac{100}{98.99.100}-\frac{98}{98.99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\left(\frac{1}{2.3}-\frac{1}{99.100}\right)\)
C = \(\frac{3}{2}.\frac{1649}{9900}\)
C = \(\frac{1649}{6600}\)
3/n . n/3 = 3.n/n.3 = 1
3/n.n/3 = 3n/3n =1