hay giup minh voi
B=1/2*x+1/2+1/6+1/12+...+1/9900=2
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MI
1
XO
14 tháng 9 2020
\(x+\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+\left(x+\frac{1}{12}\right)+...+\left(x+\frac{1}{9900}\right)=2\)
=> \(x+\left(x+\frac{1}{1.2}\right)+\left(x+\frac{1}{2.3}\right)+\left(x+\frac{1}{3.4}\right)+...+\left(x+\frac{1}{99.100}\right)=2\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=2\)(100 hạng tử x)
=> \(100x+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=2\)
=> \(100x+1-\frac{1}{100}=2\)
=> \(100x+\frac{99}{100}=2\)
=> \(100x=\frac{101}{100}\)
=> \(x=\frac{101}{10000}\)
LT
3
25 tháng 6 2016
\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)...\left(1-\frac{2}{9900}\right)\)
\(=\frac{4}{6}.\frac{10}{12}...\frac{9898}{9900}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{98.101}{99.100}\)
\(=\frac{1.2...98}{3.4...100}.\frac{4.5...101}{2.3...99}\)
\(=\frac{2}{99.100}.\frac{100.101}{2.3}\)
\(=\frac{101}{99.3}\)
\(=\frac{101}{297}\)
NT
2
HT
8 tháng 7 2016
\(x-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-...-\frac{1}{9900}=7,5\)
\(x-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}\right)=7,5\)
\(x-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)=7,5\)
\(x-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)=7,5\)
\(x-\left(1-\frac{1}{100}\right)=7,5\)
\(x-\frac{99}{100}=7,5\)
\(\Rightarrow x=7,5+\frac{99}{100}=\frac{750}{100}+\frac{99}{100}\)
\(\Rightarrow x=\frac{849}{100}=8,49\)
Dấu "." là dấu nhân nha bạn
28 tháng 6 2015
1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100
= 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101
= 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101
= 1/1 - 1/101
= 100 /101
HT
28 tháng 6 2015
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+.....+\frac{1}{9900}+\frac{1}{10100}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{99.100}+\frac{1}{100.101}\)
=\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
=\(1-\frac{1}{101}\)
=\(\frac{100}{101}\)
28 tháng 6 2015
1/2 + 1/6 + 1/12 + ... + 1/9900 + 1/10100
= 1/1.2 + 1/2.3 + 1/3.4 +... +1/99.100 + 1/100.101
= 1/1 - 1/2 + 1/2 + 1/3 - 1/3 + 1/4 +... + 1/99 - 1 / 100 + 1/100 - 1/101
= 1/1 - 1/101
= 100 /101
10 tháng 9 2020
Ta có: \(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{9900}+\frac{1}{10100}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}+\frac{1}{100.101}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
B=\(\frac{1}{2.x}+\left(\frac{1}{1.2}\frac{1}{2.3}\frac{1}{3.4}...\frac{1}{99.100}\right)\)
=\(\frac{1}{2.x}+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)\(=2\)
=\(\frac{1}{2.x}+\left(1-\frac{1}{100}\right)\)\(=2\)
=\(\frac{1}{2.x}+\frac{99}{100}\)\(=2\)
=\(\frac{1}{2.x}=2-\frac{99}{100}\)
=\(\frac{1}{2.x}=\frac{101}{200}\)
=\(2.x=200\)
=\(x=200:2=100\)
1/2 * x + 1/2 + 1/6 + 1/12 + .... + 1/9900 = 2
<=> 1/2 * x + ( 1/2 + 1/6 + 1/12 + ... + 1/9900 ) = 2
<=> 1/2 * x + ( 1 /1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 ) = 2
<=> 1/2 * x + ( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + .... + 1/99 - 1/100 ) = 2
<=> 1/2 * x + ( 1 - 1/100 ) = 2
<=> 1/2 * x + ( 100/100 - 1/100 ) = 2
<=> 1/2 * x + 99/100 = 2
<=> 1/2 * x = 2 - 99/100
<=> 1/2 * x = 101/100
<=> x = 101/100 : 1/2
<=> x = 101/100 * 2
<=> x = 101/50
Vậy x = 101/50