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3) Ta có : \(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
4)
A = \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}\right)+\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}\right)+\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}\right)+...+\frac{1}{2}.\left(\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
A = \(\frac{1}{2}.\left(1-\frac{1}{101}\right)\)
\(A=\frac{1}{2}.\frac{100}{101}\)
A = \(\frac{50}{101}\)
2, đặt tên biểu thức trên là A. Ta có :
\(A=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{10100}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{100.101}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{100}-\frac{1}{101}\)
\(A=1-\frac{1}{101}\)
\(A=\frac{100}{101}\)
1) \(\frac{1}{1}.\frac{1}{2}+\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{5}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}\)
\(=1-\frac{1}{5}\)
\(=\frac{4}{5}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{1017.2019}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2018}{2019}\)
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+....+\frac{2}{2017\cdot2019}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(=1-\frac{1}{2019}=\frac{2018}{2019}\)
Bài 1 :
\(S=1.3+3.5+5.7+...+99.101=3+15+35+...9999\)
Ta thấy :
\(3=2^2-1\)
\(15=4^2-1\)
\(35=6^2-1\)
.....
\(9999=100^2-1\)
\(\Rightarrow S=2^2+4^2+...+100^2-\left(1\right).\left(\left(100-2\right):2+1\right)\)
\(\Rightarrow S=\dfrac{100.\left(100+1\right)\left(2.100+1\right)}{6}-51\)
\(\Rightarrow S=\dfrac{100.101.201}{6}-51=338299\)
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 = \(\frac{18}{19}x5=\frac{90}{19}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+\frac{1}{11}\)
\(=\frac{1}{1}-\frac{1}{11}\)
\(=\frac{10}{11}\)
Bạn gõ lại đề đi :v
Đọc chả hiểu đề gì cả ... đề k có x
Mà phía dưới có cái đáp số x= ... là sao ??
a)(\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{11.12}\)). x=\(\frac{1}{3}\)
(1-\(\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{11}_{ }+\frac{1}{12}\)).x=\(\frac{1}{3}\)
(1+\(\frac{1}{12}\)).x=\(\frac{1}{3}\)
x=\(\frac{1}{3}:\frac{13}{12}\)
x=\(\frac{4}{13}\)
https://invite.duolingo.com/BDHTZTB5CWWKSFQDXLOOOK3AAA
a: \(A=2\cdot1+4\cdot2+6\cdot3+8\cdot4+\cdots+40\cdot20\)
\(=2\cdot1+2\cdot1\cdot2^2+2\cdot1\cdot3^2+\cdots+2\cdot1\cdot20^2\)
\(=2\cdot1\cdot\left(1^2+2^2+\ldots+20^2\right)\)
\(=2\cdot\frac{20\cdot\left(20+1\right)\left(2\cdot20+1\right)}{6}\)
\(=\frac{40\cdot21\cdot41}{6}=20\cdot7\cdot41=140\cdot41=5740\)
b: \(B=1\cdot5+2\cdot6+\cdots+50\cdot54\)
\(=1\left(1+4\right)+2\left(2+4\right)+\cdots+50\left(2+4\right)\)
\(=\left(1^2+2^2+\cdots+50^2\right)+4\left(1+2+\cdots+50\right)\)
\(=\frac{50\cdot\left(50+1\right)\left(2\cdot50+1\right)}{6}+4\cdot\frac{50\cdot51}{2}\)
\(=\frac{50\cdot51\cdot101}{6}+2\cdot50\cdot51\)
\(=25\cdot17\cdot101+100\cdot51=5100+42925=48025\)
c: \(C=1\cdot6+2\cdot7+\cdots+50\cdot55\)
\(=1\left(1+5\right)+2\left(2+5\right)+\cdots+50\left(50+5\right)\)
\(=\left(1^2+2^2+\cdots+50^2\right)+5\left(1+2+\cdots+50\right)\)
\(=\frac{50\cdot\left(50+1\right)\left(2\cdot50+1\right)}{6}+5\cdot50\cdot\frac{51}{2}\)
\(=\frac{50\cdot51\cdot101}{6}+5\cdot25\cdot51\)
\(=25\cdot17\cdot101+5\cdot25\cdot51\)
\(=25\left(17\cdot101+5\cdot51\right)=25\cdot1972=49300\)