Bài 2:Tính tổng
a) \(\frac{1}{2x3}+\frac{1}{3x4}+....+\frac{1}{49x50}\)
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NT
5
9 tháng 5 2019
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{49.50}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{2}+\left(\frac{1}{3}-\frac{1}{3}\right)+\left(\frac{1}{4}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+...+\left(\frac{1}{49}-\frac{1}{49}\right)-\frac{1}{50}\)
\(=\frac{1}{2}-\frac{1}{50}=\frac{12}{25}\)
~ Hok tốt ~
NH
9 tháng 5 2019
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{49.50}\)
\(=2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)
\(=2\left(\frac{1}{2}-\frac{1}{50}\right)=2.\frac{12}{25}=\frac{24}{25}\)
7 tháng 4 2016
1/1x2+1/2x3+...+1/49x50
=1-1/2+1/2-1/3+.....+1/49-1/50
=1-1/50(1)
Ta co 1(2)
So sanh (1) voi (2) ta thay 1-1/50<1
=>1/1x2+...+1/49x50<1
(Phuong phap khu)
7 tháng 4 2016
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)
Vậy \(\frac{49}{50}<1\)
S
2
TP
9 tháng 8 2017
Ta có:
\(\frac{1}{3.4}.x+\frac{1}{4.5}.x+...+\frac{1}{49.50}.x=1\)
\(\Rightarrow x.\left(\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)=1\)
\(\Rightarrow x.\left(\frac{1}{3}-\frac{1}{50}\right)=1\Leftrightarrow x.\frac{47}{150}=1\)
\(\Rightarrow x=1:\frac{47}{150}\Leftrightarrow x=\frac{150}{47}\)
BK
1
11 tháng 9 2018
\(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{19\cdot20}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{2}-\frac{1}{20}\)
\(=\frac{9}{20}\)
11 tháng 9 2018
\(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ \(\frac{1}{4x5}\)+ ... + \(\frac{1}{18x19}\)+ \(\frac{1}{19x20}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{5}\)+ ... + \(\frac{1}{18}\)- \(\frac{1}{19}\)+ \(\frac{1}{19}\)- \(\frac{1}{20}\)
= \(\frac{1}{2}\)- \(\frac{1}{20}\)
= \(\frac{18}{40}\)= \(\frac{9}{20}\)
CT
3
H
29 tháng 4 2019
\(1+\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2017.2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2018}{2018}-\frac{1}{2018}\right)\)
\(=1+\left(\frac{2017}{2018}\right)\)
\(=\frac{2018}{2018}+\frac{2017}{2018}=\frac{4035}{2018}\)
L
1 tháng 5 2019
\(1+\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}...+\frac{1}{2017\cdot2018}\)
\(=1+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}...+\frac{1}{2017}-\frac{1}{2018}\right)\)
\(=1+\left(1-\frac{1}{2018}\right)\)
\(=1+\frac{2017}{2018}\)
\(=1+\frac{2017}{2018}\)
\(=\frac{4035}{2018}\)
=1/2-1/3+1/3-1/4+.....+1/49+1/50
=1/2-1/50
=25/50-1/50
=24/50
=12/25
\(\frac{1}{2x3}\)+ \(\frac{1}{3x4}\)+ ... + \(\frac{1}{49x50}\)
= \(\frac{1}{2}\)- \(\frac{1}{3}\)+ \(\frac{1}{3}\)- \(\frac{1}{4}\)+ ... + \(\frac{1}{49}\)- \(\frac{1}{50}\)
= \(\frac{1}{2}\)- \(\frac{1}{50}\)
=\(\frac{12}{25}\)