Tính giá trị biểu thức:
\(\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+....+\frac{324}{17.19}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
S = \(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+\frac{5}{7.9}+.......+\frac{5}{17.19}\)
S : 5 = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+.....+\frac{1}{17.19}\)
S : 5 = \(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}+.......+\frac{1}{17}-\frac{1}{19}\)
=> S : 5 = 1 - \(\frac{1}{19}=\frac{19}{19}-\frac{1}{19}=\frac{18}{19}\)
=> S = \(\frac{18}{19}x5=\frac{90}{19}\)
Ta có:
\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{91.93}+\frac{5}{93.95}=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{91.93}+\frac{1}{93.95}\right)=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{91.93}+\frac{2}{93.95}\right)\)
\(\Rightarrow A=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{93}+\frac{1}{93}-\frac{1}{95}\right)=\frac{5}{2}\left(1-\frac{1}{95}\right)=\frac{5}{2}.\frac{94}{95}=\frac{47}{19}\)
Vậy \(A=\frac{47}{19}\)
\(A=\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{93.95}\)
\(A=5\cdot\frac{1}{2}\cdot\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-....-\frac{1}{95}\right)\)
\(A=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{95}\right)=\frac{5}{2}\cdot\frac{94}{95}=\frac{47}{19}\)
y=2/1.3 + 2/3.5 +2/5.7 +...+2/99.101
y= 1.(1-1/3+1/3-1/5+1/5-1/7+....+1/99-1/101)
y=1. ( 1-1/101)
y= 1. 100/101
y=100/101
\(A=\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{1006^2}{2011.2013}\)
\(\Leftrightarrow4A=\frac{2^2.1^2}{2^2-1}+\frac{2^2.2^2}{4^2-1}+...+\frac{2^2.1006^2}{2012^2-1}\)
\(=1006+\left(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2011.2013}\right)\)
\(=1006+\frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(=1006+\frac{1}{2}\left(1-\frac{1}{2013}\right)=\frac{2026084}{2013}\)
\(\Rightarrow A=\frac{506521}{2013}\)
\(\frac{2}{5}A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{93.95}\)
\(\frac{2}{5}A=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{93}-\frac{1}{95}\)
\(\frac{2}{5}A=1-\frac{1}{95}=\frac{94}{95}\)
\(A=\frac{47}{19}\)
A= 5/1.3 +5/3.5 +5/5.7+....+5/91.93 + 5/93.95
A= 5/2 . (1-1/3+1/3-1/5+1/5-1/7+...+1/91-1/93+1/93-1/95)
A=5/2. (1-1/95)
A= 5/2 . 94/95
A=47/19
Nhớ k nha
c) \(C=1.2+2.3+3.4+...+98.99\)
\(\Rightarrow3C=1.2\left(3-0\right)+2.3\left(4-1\right)+3.4\left(5-2\right)+...+98.99\left(100-97\right)\)
\(=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+98.99.100-97.98.99\)
\(=98.99.100\)
\(\Rightarrow C=\frac{98.99.100}{3}=323400\)
d) \(D=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
=\(\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{2.4}+...+\frac{2}{8.10}\right)\)
= \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+....+\frac{1}{8}-\frac{1}{10}\right)\)
= \(\frac{1}{2}.\left(1+\frac{1}{2}-\frac{1}{9}-\frac{1}{10}\right)\)
=\(\frac{29}{45}\)
\(\frac{4}{1.3}+\frac{16}{3.5}+\frac{36}{5.7}+...+\frac{324}{17.19}\)
\(=\frac{2^2}{1.3}+\frac{4^2}{3.5}+\frac{6^2}{5.7}+...+\frac{18^2}{17.19}\)
\(=2\left(\frac{2.1^2}{1.3}+\frac{2.2^2}{3.5}+\frac{2.3^2}{5.7}+...+\frac{2.9^2}{17.19}\right)\)
\(=2\left(\frac{1}{1}-\frac{1}{3}+\frac{2^2}{3}-\frac{2^2}{5}+\frac{3^2}{5}-\frac{3^2}{7}+...+\frac{9^2}{17}-\frac{9^2}{19}\right)\)
\(=2\left[1+\frac{2^2-1}{3}+\frac{3^2-2^2}{5}+...+\frac{9^2-8^2}{17}-\frac{9^2}{19}\right]\)
\(=2\left(1+1+1....+1-\frac{81}{19}\right)=2\left(9-\frac{81}{19}\right)=\frac{180}{19}\)