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HT
3
23 tháng 6 2017
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..................+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+..................+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+.......+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{99}-\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+.......+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.......+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+.......+\dfrac{1}{100}\)
13 tháng 6 2017
Bài này đăng nhiều lắm rồi đó bạn: Câu hỏi của Trieu Nguyen - Toán lớp 6 | Học trực tuyến
L
1
6 tháng 1
\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)+\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=\left(1-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{99}-\dfrac{1}{100}\right)\)
=\(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)
\(\left(\dfrac{2}{1\cdot2}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)\)=>\(2\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\cdot\dfrac{x^2+x+1945}{2}>1975\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\right)\)
=>\(x^2+x+1945>1975\)
=>\(x^2+x-30>0\)
=>(x+6)(x-5)>0
TH1: \(\left\{{}\begin{matrix}x+6>0\\x-5>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>-6\\x>5\end{matrix}\right.\)
=>x>5
TH2: \(\left\{{}\begin{matrix}x+6< 0\\x-5< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< -6\\x< 5\end{matrix}\right.\)
=>x<-6
S
30 tháng 7 2018
\(=\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)-\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{99.100.101}\right)\)
\(=\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{99.100}-\frac{1}{100.101}\right)\)
\(=\left(1-\frac{1}{100}\right)-\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{100.101}\right)\)
\(=\frac{99}{100}-\frac{1}{2}\cdot\frac{5049}{10100}=\frac{99}{100}-\frac{5049}{20200}=\frac{14949}{20200}\)
LM
13 tháng 8 2016
A=\(\frac{1}{1.2}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{2017.2018}\)
A=1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+...+\(\frac{1}{2017}\)-\(\frac{1}{2018}\)
A=1-\(\frac{1}{2018}\)
A=\(\frac{2017}{2018}\)
13 tháng 8 2016
\(A=\frac{1}{1.2}+\frac{1}{3.4}+\frac{1}{5.6}+...+\frac{1}{2017.2018}\)
\(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(A=\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2018}\right)\)
\(A=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{2017}+\frac{1}{2018}\right)-\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{1009}\right)\)
\(A=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2017}+\frac{1}{2018}\)
Đến đây bình thường ta nhóm 2 số vào với nhau nhưng ở đây có lẻ số hạng nên không nhóm được => đề sai
15 tháng 12 2017
\(B=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=1-\frac{1}{100}\)
\(B=\frac{99}{100}\)
= 22.50.51.52 : 6 – 51.50 = 88400 – 2550 = 85850.
\(A=\left(2-1\right).2+\left(4-1\right).4+\left(6-1\right).6+...+\left(100-1\right).100\\ A=2^2-2+4^2-4+6^2-6+...+100^2-100\\ A=\left(2^2+4^2+...+100^2\right)-\left(2+4+...+100\right)\\ A=2^2\left(1+2^2+3^2+...+50^2\right)-\dfrac{\left(100+2\right).50}{2}\\ A=\dfrac{4.50.51.52}{6}-\dfrac{102.50}{2}=85850\)