Tính giá trị biểu thức:
A=\(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+...+\frac{1004^2}{2007.2009}+\frac{1005^2}{2009.2011}+\frac{1006^2}{2011.2013}\)
Những câu hỏi liên quan
bai 1: Tinh gia tri cuar cac bieu thuc sau
a, B=\(\frac{3}{0,29972997...}+\frac{3}{0,29972997...}+\frac{3}{0,0029972997....}\)
b, C=\(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+....+\frac{1005^2}{2009.2011}_{+\frac{1006^2}{2011.2013}}\)
cach giai voi nha
Tính S=\(\frac{1^2}{1.3}+\frac{2^2}{3.5}+\frac{3^2}{5.7}+....+\frac{1004^2}{2007.2009}\)
mình đang cần gấp bạn nào giải được chi tiết mình tặng 3 tick vì có 3 nick
Tính \(\frac{1^2}{1.3}+\frac{2^2}{3.5}+...+\frac{2005^2}{2009.2011}+\frac{2006^2}{2011.2013}\)
Bài 2 Tính:
A=\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{2011.2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
Đúng 0
Bình luận (0)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2011.2013}\)
\(A=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2011}-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2013}\right)\)
\(A=\frac{1}{2}.\frac{2012}{2013}\)
\(A=\frac{1006}{2013}\)
Đúng 0
Bình luận (0)
A = 1/1. x 1/3 + 1/3 x 1/5 + 1/5 x 1/7 x.....+1/2011 x 1/2013
A=1/1 x 1/2013(bạn triệt tiêu 1/3 ,1/5 ,1/7,1/2011 nha )
A =1/2013
Đúng 0
Bình luận (0)
Tính P = \(\dfrac{1^2}{3.5}+\dfrac{2^2}{3.5}+\dfrac{3^2}{5.7}+....+\dfrac{1004^2}{2007.2009}+\dfrac{1005^2}{2009.2011}\)
\(P=\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+...+\dfrac{1005^2}{2009.2011}\)
\(\Leftrightarrow4P=\dfrac{4.1^2}{1.3}+\dfrac{4.2^2}{3.5}+...+\dfrac{4.1005^2}{2009.2011}\)
\(=\dfrac{2^2}{2^2-1}+\dfrac{4^2}{4^2-1}+...+\dfrac{2010^2}{2010^2-1}\)
\(=2009+\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2009.2011}\right)\)
\(=2009+\dfrac{1}{2}\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=2009+\dfrac{1}{2}\left(1-\dfrac{1}{2011}\right)=2009+\dfrac{1005}{2011}\)
Đúng 0
Bình luận (0)
Ace Legona Akai Haruma Phương AnPhương AnVõ Đông Anh Tuấn làm jum Hung nguyen
Đúng 0
Bình luận (1)
\(y=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
Tính giá trị biểu thức trên
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
Đúng 0
Bình luận (0)
Tính giá trị của biểu thức:Tfrac{4}{2.4}+frac{4}{4.6}+frac{4}{6.8}+...+frac{4}{2008.2010}Afrac{1}{1.3}+frac{1}{3.5}+frac{1}{5.7}+...+frac{1}{2007.2009}+frac{1}{2009.2011}Cleft(1-frac{1}{2}right)left(1-frac{1}{3}right)left(1-frac{1}{4}right)...left(1-frac{1}{1000}right)Sfrac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}Bleft(1+frac{1}{2}right)left(1+frac{1}{3}right)left(1+frac{1}{4}right)...left(1+frac{1}{100}right)Dleft(1-frac{1}{17}right)left(1-frac{2}{17}right)left(1-frac{3}{17}right)...left(1-f...
Đọc tiếp
Tính giá trị của biểu thức:
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009.2011}\)
\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{1000}\right)\)
\(S=\frac{1+2+2^2+2^3+...+2^{2008}}{1-2^{2009}}\)
\(B=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\left(1+\frac{1}{4}\right)...\left(1+\frac{1}{100}\right)\)
\(D=\left(1-\frac{1}{17}\right)\left(1-\frac{2}{17}\right)\left(1-\frac{3}{17}\right)...\left(1-\frac{27}{17}\right)\)
\(T=\frac{4}{2.4}+\frac{4}{4.6}+\frac{4}{6.8}+...+\frac{4}{2008.2010}\)
\(T=2.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{2008.2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(T=2.\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(T=2.\frac{502}{1005}=\frac{1004}{1005}\)
\(\Rightarrow T=\frac{1004}{1005}\)
Đúng 0
Bình luận (0)
\(A=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2007.2009}+\frac{1}{2009+2011}\)
\(A=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{2009+2011}\right)\)
\(A=\frac{1}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
\(A=\frac{1}{2}.\frac{2010}{2011}\)
\(\Rightarrow A=\frac{1005}{2011}\)
Đúng 0
Bình luận (0)
\(C=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{100}\right)\)
\(C=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{99}{100}\)
\(C=\frac{1.2.3...99}{2.3.4...100}\)
\(\Rightarrow C=\frac{1}{100}\)
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
tính giá trị của biểu thức
a) A=\(\frac{1}{1.2}\) + \(\frac{1}{2.3}\) + \(\frac{1}{3.4}\) + \(\frac{1}{4.5}\) + ...+\(\frac{1}{99.100}\)
b) B= \(\frac{2}{1.3}\)+\(\frac{2}{3.5}\) + \(\frac{2}{5.7}\)+\(\frac{2}{7.9}\) +...+\(\frac{2}{97.99}\)
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
b) \(B=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\right)\)
\(=2.\left(1-\frac{1}{99}\right)\)
\(=2.\frac{98}{99}\)
\(=\frac{196}{99}=1\frac{97}{99}\)
Đúng 0
Bình luận (1)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(B=\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{97.99}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\)
\(=1-\frac{1}{99}\)
\(=\frac{98}{99}\)
Đúng 0
Bình luận (3)
A=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{99.100}\)
=>\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=>\(\frac{1}{1}-\frac{1}{100}\)
=>\(\frac{99}{100}\)
B=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{97.99}\)
=>\(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+....+\frac{1}{97}-\frac{1}{99}\)
=>\(\frac{1}{1}-\frac{1}{99}\)
=>\(\frac{98}{99}\)
Đúng 0
Bình luận (0)
Tính
a)S1=\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
b)S2=\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
c)S3=\(\frac{1}{10.9}+\frac{1}{18.13}+\frac{1}{26.17}+...+\frac{1}{802.405}\)
\(S1=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+....+\frac{2}{99.101}\)
\(S1=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....-\frac{1}{101}=\frac{1}{1}-\frac{1}{101}=\frac{100}{101}\)
\(S2=\frac{5}{1.3}+\frac{5}{3.5}+....+\frac{5}{99.101}\)
\(S2=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-.....-\frac{1}{101}\right)=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)=\frac{5}{2}\cdot\frac{100}{101}=\frac{250}{101}\)
Đúng 0
Bình luận (0)