Ta có:B = \(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}......\dfrac{98.100}{99^2}\)
\(=\dfrac{1.2.3......98}{2.3.4......99}.\dfrac{3.4.5.....100}{2.3.4.....99}=\dfrac{1}{99}.\dfrac{100}{2}=\dfrac{100}{198}\)
Vậy B = \(\dfrac{100}{198}\)
câu 1 tính
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)....\left(1+\dfrac{1}{2015.2017}\right)\)
1, P = \(\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}\) - \(\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{2}{2004}}\)
2, Q = ( \(\dfrac{1,5+1-0,75}{2,5+\dfrac{5}{3}-1,25}\) + \(\dfrac{0,375-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{-0,625+0,5-\dfrac{5}{11}-\dfrac{5}{12}}\) ) : \(\dfrac{1980}{3758}\) + 155
3, A = 1.3 + 2.4 + 3.5 +....+ 97.99 + 98.100
4, B = 1.2.3 + 2.3.4. +...+ 48.49.50
5, C = \(\dfrac{1}{1.2.3.4}\) + \(\dfrac{1}{2.3.4.5}\) +...+ \(\dfrac{1}{27.28.29.30}\)
6, D = 1 + \(2^2\) + \(2^4\) + \(2^6\) + .... +\(2^{200}\)
7, E = \(\dfrac{1}{3.5}\)+ \(\dfrac{5}{5.7}\) +...+ \(\dfrac{1}{97.99}\)
\(A=\dfrac{1}{2}\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2015.2017}\right)\)
HELP ME !!! THANK
Tìm x nguyên biết:
a)\(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{\left(2x-1\right).\left(2x+1\right)}=\dfrac{49}{99}\)
b)\(1-3+3^2-3^3+...+\left(-3\right)^x=\dfrac{9^{1006}-1}{2}\)
\(\dfrac{1^2}{1.3}+\dfrac{2^2}{3.5}+\dfrac{3^2}{5.7}\)
Tính:
a,A=\(\dfrac{12^{15}.3^4-4^5.3^9}{27^3.2^{10}-32^3.3^9}\)
b. B= \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^3.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{99}{49^2.50^2}\)
Tính tổng
\(S=\dfrac{1}{1.3}-\dfrac{1}{2.4}+\dfrac{1}{3.5}-\dfrac{1}{4.6}+\dfrac{1}{5.7}-\dfrac{1}{6.8}+\dfrac{1}{7.9}-\dfrac{1}{8.10}\)
a) Tìm x biết: \(\dfrac{x+1}{100}+\dfrac{x+2}{99}=\dfrac{x+3}{98}+\dfrac{x+4}{97}\)
b) So sánh \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\) với 1
c) Tìm GTNN của: A= |x-10|+|x-5|
Tính: \(S=\left(1+\dfrac{1}{1.3}\right).\left(1+\dfrac{1}{2.4}\right).\left(1+\dfrac{1}{3.5}\right)...\left(1+\dfrac{1}{2016.2018}\right)\)
