Ta có: \(S=\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+...+\dfrac{199}{99^2.100^2}\)
\(\Rightarrow S=\dfrac{1}{1^2}-\dfrac{1}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{3^2}+...+\dfrac{1}{99^2}-\dfrac{1}{100^2}\)
\(\Rightarrow S=1-\dfrac{1}{100^2}\)
\(\Rightarrow S=1-\dfrac{1}{10000}=\dfrac{9999}{10000}\)
Bài 1: Cho D= \(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+\dfrac{7}{3^2.4^2}+...+\dfrac{19}{9^2.10^2}\). Chứng minh D < 1
cho A=\(\dfrac{3}{1^2.2^2}+\dfrac{5}{2^2.3^2}+........+\dfrac{2n-1}{\left(n-1\right)^2.n^2}\)
CMR:A<1
*Rút gọn
1) G=\(\dfrac{2}{3}+\dfrac{2}{3^3}+\dfrac{2}{3^5}+...+\dfrac{2}{3^{99}}\)
2) H=\(\dfrac{1}{2}-\dfrac{1}{2^4}+\dfrac{1}{2^7}-\dfrac{1}{2^{16}}+...-\dfrac{1}{2^{58}}\)
3) E=\(\dfrac{-1}{3}+\left(\dfrac{-1}{3}\right)^2+\left(\dfrac{-1}{3}\right)^3+...+\left(\dfrac{-1}{100}\right)^{100}\)
Tính tổng: \(S=\dfrac{3}{1^2.3}+\dfrac{5}{\left(1^2+2^2\right).4}+\dfrac{7}{\left(1^2+2^2+3^2\right).5}+...+\dfrac{2n+1}{\left(1^2+2^2+3^2+...+n^2\right).\left(n+2\right)}\)
(mấy chế làm giùm đi , nhanh nhanh giùm nhen,mai học rồi)
10 Rút gọn:
a) A= 1+2+22+23+24+...+249+250
b) B= \(\dfrac{1}{2}+(\dfrac{1}{2})^2+(\dfrac{1}{2})^3+(\dfrac{1}{2})^4+(\dfrac{1}{2})^5+...+(\dfrac{1}{2})^{99}+(\dfrac{1}{2})^{100}\)
\(A=\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+.+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}.\)So sánh A với 2
1. Tính:
a. \(\dfrac{-1}{4}+\dfrac{5}{6}\) b. \(\dfrac{5}{12}+\dfrac{-7}{8}\) c. \(\dfrac{-7}{6}+\dfrac{-3}{10}\) d. \(\dfrac{-3}{7}+\dfrac{5}{6}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}\) b. \(\dfrac{-13}{12}-\dfrac{5}{18}\) c. \(\dfrac{-2}{5}-\dfrac{-3}{11}\) d. \(0,6--1\dfrac{2}{3}\)
3. Tính :
a. \(\dfrac{-1}{39}+\dfrac{-1}{52}\) b. \(\dfrac{-6}{9}-\dfrac{12}{16}\) c. \(\dfrac{-3}{7}-\dfrac{-2}{11}\) d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
So sánh A và B với \(\dfrac{1}{2}\) biết :
\(A=\dfrac{1}{1.2^2}+\dfrac{1}{2.3^2}+\dfrac{1}{3.4^2}+........+\dfrac{1}{49.50^2}\) và
\(B=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+.......+\dfrac{1}{50^2}\)
Tính:
a/\(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{3+\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{3}{4}}{2-\dfrac{2}{2}+\dfrac{2}{3}-\dfrac{2}{4}}\)
b/\(\dfrac{1+\dfrac{1}{4}+\dfrac{1}{1+\dfrac{1}{4}}}{1-\dfrac{1}{4}-\dfrac{1}{1-\dfrac{1}{4}}}\)
c/\(\dfrac{\dfrac{2}{5}-\dfrac{7}{5}}{\dfrac{2}{5}-\dfrac{\dfrac{3}{4}}{\dfrac{3}{4}.\dfrac{3}{7}-1}}-\dfrac{1}{\dfrac{3}{7}\left(\dfrac{3}{4}.\dfrac{3}{7}.\dfrac{2}{5}-\dfrac{2}{5}-\dfrac{3}{4}\right)}\)
d/\(\left(\dfrac{\dfrac{4}{3}}{2+\dfrac{4}{3}}+\dfrac{2-\dfrac{4}{3}}{\dfrac{4}{3}}\right).\left(\dfrac{\dfrac{2}{3}}{4+\dfrac{2}{3}}-\dfrac{4-\dfrac{2}{3}}{\dfrac{2}{3}}\right)\)
Giúp mik với các bạn ơi 1 bài thôi cug đc.
Câu 1)
1) \(\dfrac{11}{24}\)−\(\dfrac{5}{41}\)+\(\dfrac{13}{24}\)+0,5−\(\dfrac{36}{41}\)=
2)12÷\(\left(\dfrac{3}{4}-\dfrac{5}{6}\right)^2\)=
3) (\(1+\dfrac{2}{3}-\dfrac{1}{4}\))\(\left(0,8-\dfrac{3}{4}\right)^2\) =
4)\(16\dfrac{2}{7}\)÷(\(\dfrac{-3}{5}\))+\(28\dfrac{2}{7}\)÷\(\dfrac{3}{5}\)
5)\(\left(2^2\div\dfrac{4}{3}-\dfrac{1}{2}\right)\times\dfrac{6}{5}-17\)
6)\(\left(\dfrac{1}{3}\right)^{50}\times\left(-9\right)^{25}-\dfrac{2}{3}\div4\)
