Tính tổng:
A= 5/1.4+5/4.7+...+5/100.103
B= 1/15+1/35+...+1/2499
K
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NT
tính giá trị biểu thức
C=1/15+1/35+.................+1/2499
B=5/1.4+5/4.7+...................5/100.103
0
AM
4 tháng 7 2015
a)\(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}=\frac{5}{3}\cdot\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{100.103}\right)\)
\(=\frac{5}{3}\cdot\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}\cdot\left(1-\frac{1}{103}\right)=\frac{5}{3}\cdot\frac{102}{103}=\frac{170}{103}\)b)\(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\right)=\frac{1}{2}\cdot\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\cdot\frac{16}{51}=\frac{8}{51}\)
5 tháng 5 2017
Câu a) bạn Ác Mộng làm rồi nên mình làm b) nha
b)Gọi A = \(\frac{1}{15}+\frac{1}{35}+...+\frac{1}{2499}\)
\(A=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\)
\(2A=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}\right)\)
\(2A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{49.51}\)
\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}\)
\(2A=\frac{1}{3}-\frac{1}{51}\)
\(2A=\frac{16}{51}\)
\(A=\frac{16}{51}:2\)
\(A=\frac{8}{51}\)
6 tháng 5 2018
A=2.(1/1.3 + 1/3.5 + 1/5.7 +.......+1/99.101)
=2.(1/1 + 1/3 + 1/5 + 1/5 + 1/7 +...+1/99 + 1/101)
=2.(1-1/101)
=2.(101/101-1/101)
=2.100/101
200/101
6 tháng 5 2018
B=2.(1/1.3+1/3.5+1/3.1+....+1/99.101)
=2.(1/1+1/3+1/3+1/5+1/3+1/7+....+1/99+1/101)
=2.(1/1+1/101)
=2.(101/101+1/101)
=2.102/101
=204/101
7 tháng 6 2016
a) \(\frac{1}{n}-\frac{1}{n+a}=\frac{\left(n+a\right)-n}{n\left(n+a\right)}=\frac{a}{a\left(n+a\right)}\) (đpcm)
b) \(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\right)=\frac{5}{3}.\left(1-\frac{1}{103}\right)=\frac{5}{3}.\frac{102}{103}=\frac{170}{103}\)
\(C=\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{49.51}=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{51}=\frac{1}{3}-\frac{1}{51}=\frac{16}{51}\)
DK
1
22 tháng 3 2018
Ta có : \(A=\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{2}+\left(-\frac{1}{3}+\frac{1}{3}\right)+\left(-\frac{1}{4}+\frac{1}{4}\right)+...+\left(-\frac{1}{99}+\frac{1}{99}\right)-\frac{1}{100}\)
\(A=\frac{1}{2}+0+0+..+0-\frac{1}{100}\)
\(A=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(B=\frac{5}{1.4}+\frac{5}{4.7}+..+\frac{5}{100.103}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
\(B=1+\left(-\frac{1}{4}+\frac{1}{4}\right)+\left(-\frac{1}{7}+\frac{1}{7}\right)+...+\left(-\frac{1}{100}+\frac{1}{100}\right)-\frac{1}{103}\)
\(B=1+0+0+...+0-\frac{1}{103}\)
\(B=1-\frac{1}{103}=\frac{102}{103}\)
So sánh : A < B vì 49/100 < 102/103 (49.103 < 102 . 100)
a) \(A=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(\Leftrightarrow A=\dfrac{5}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}\left(1-\dfrac{1}{103}\right)\)
\(\Leftrightarrow\dfrac{5}{3}.\dfrac{102}{103}\)
\(\Leftrightarrow\) \(A=\dfrac{170}{103}\)
b) \(B=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(B=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(B=\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{51}\right)\)
\(B=\dfrac{1}{2}.\dfrac{16}{51}\)
\(B=\dfrac{8}{51}\)
A = \(\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
A = \(\dfrac{5}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
A = \(\dfrac{5}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{100}+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\left(\dfrac{1}{4}-\dfrac{1}{4}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{100}-\dfrac{1}{100}\right)-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-0-0-...-0-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{1}{1}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\left[\dfrac{103}{103}-\dfrac{1}{103}\right]\)
A = \(\dfrac{5}{3}.\dfrac{102}{103}\)
A = \(\dfrac{170}{103}\)
B = \(\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
B = \(\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
B = \(\dfrac{1}{2}.\left(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\right)\)
B = \(\dfrac{1}{2}.\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-...-\dfrac{1}{49}+\dfrac{1}{49}-\dfrac{1}{51}\right)\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{49}-\dfrac{1}{49}\right)-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-0-0-...-0-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{1}{3}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\left[\dfrac{17}{51}-\dfrac{1}{51}\right]\)
B = \(\dfrac{1}{2}.\dfrac{16}{51}\)
B = \(\dfrac{8}{51}\)