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DH
4
T
21 tháng 6 2019
#)Giải :
\(\frac{1}{3}+\frac{1}{15}+\frac{1}{35}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
\(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{y\left(y+2\right)}=\frac{50}{101}\)
\(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{y}-\frac{1}{y+2}=\frac{50}{101}\)
\(1-\frac{1}{y+2}=\frac{50}{101}\)
\(\Leftrightarrow\frac{1}{y+2}=\frac{51}{101}\)
\(\Leftrightarrow y+2=\frac{101}{51}\)
\(\Leftrightarrow x=-\frac{1}{51}\)
VQ
2
13 tháng 1 2017
2/3 + 2/15 + 2/35 + ... + 2/n = 322/323
2/1x3 + 2/3x5 + 3/5x7 + ... + 2/nx(n+2) = 322/323
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .... + 1/n-1 - 1/(n+2) = 322/323
1 - 1/n+2 = 322/323
1/n+2 = 1 - 322/323
1/n+2 = 1/323
=> x + 2 = 323
n = 323 - 2 = 321
13 tháng 1 2017
2/3 + 2/15 + 2/35 + ... + 2/n = 322/323
2/1x3 + 2/3x5 + 3/5x7 + ... + 2/nx(n+2) = 322/323
1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + .... + 1/n-1 - 1/(n+2) = 322/323
1 - 1/n+2 = 322/323
1/n+2 = 1 - 322/323
1/n+2 = 1/323
=> x + 2 = 323
n = 323 - 2 = 321
DT
1
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
23 tháng 3 2023
( \(\dfrac{2}{15}\) + \(\dfrac{2}{35}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{2\times7}{15\times7}\) + \(\dfrac{2\times3}{35\times3}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{14}{105}\) + \(\dfrac{6}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{20}{105}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
( \(\dfrac{4}{21}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
(\(\dfrac{12}{63}\) + \(\dfrac{2}{63}\)) : \(x\) = \(\dfrac{1}{18}\)
\(\dfrac{2}{9}\) : \(x\) = \(\dfrac{1}{18}\)
\(x\) = \(\dfrac{2}{9}\) : \(\dfrac{1}{18}\)
\(x\) = 4
24 tháng 4 2021
\(\frac{2}{3}\cdot y-\frac{12}{3}:\left(\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}+\frac{2}{143}\right)=\frac{1}{3}\)\(\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+\frac{2}{9\cdot11}+\frac{2}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+\frac{9-7}{7\cdot9}+\frac{11-9}{9\cdot11}+\frac{13-11}{11\cdot13}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(1+\frac{1}{3}-\frac{1}{3}+\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+\frac{1}{9}-\frac{1}{9}+\frac{1}{11}-\frac{1}{11}+\frac{1}{13}\right)\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\left(\frac{1}{1}+\frac{1}{3}\right)=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4:\frac{4}{3}\)\(=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-4\cdot\frac{3}{4}=\frac{1}{3}\)
\(\frac{2}{3}\cdot y-3=\frac{1}{3}\)
\(\frac{2}{3}\cdot y=\frac{1}{3}+3\)
\(\frac{2}{3}\cdot y=\frac{10}{3}\)
\(y=\frac{10}{3}:\frac{2}{3}\)
y=5
=> 2/1x3 +2/3x5+2/5x7+2/7x9+...+2/nx(n+2)
=>1-1/3+1/3-1/5+1/5-1/7+1/7-1/9+1/9+...+1/n-1/n+2
=>1-1/n+2=100/101
1/n+2=1-100/101
1/n+2=1/101
=>n+2=101
=>n=101-2
=>n=99