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\(=\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n.\left(n+1\right)}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{n.\left(n+1\right)}\)
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2001}\)
\(2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
\(2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2001}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4002}\)
\(\frac{1}{n+1}=\frac{1}{2001}\)
=>n+1=2001
=>n=2000
\(A=3-\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-\frac{1}{20}-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}\)
\(A=3-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(A=3-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(A=3-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(A=3-\left(1-\frac{1}{8}\right)\)
\(A=3-\frac{5}{8}\)
\(A=\frac{19}{8}\)
a) để 5/n-1 là số nguyên thì 5 chia hết cho n-1
=> n-1 thuộc Ư(5)=( 1, -1, 5, -5)
ta có
n-1=1=>n=2
n-1=-1=>n=0
n-1=5=>n=6
n-1=-5=>n=-4
mà n là số tự nhiên => n thuộc 2,0,6
máy mik bị lỗi bàn phím nên phải gõ ngoặc khác thay thế TvT, sorry nghen
b) M=(1-1000/2016) *...*(1-2016/2016)*(1-2017/2016)
=>M=(1-1000/2016)*.....*0*(1-2017/2016)
=>M=0
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}\)
\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Rightarrow2\cdot\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}\div2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{4032}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4032}\)
\(\Rightarrow x+1=4032\Rightarrow x=4031\)
Vậy \(x=4031\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.x+1}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2032}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{2032}\)
=> \(\frac{1}{x+1}=\frac{1}{2032}\)
Vì 1 = 1
=> x + 1 = 2032
=> x = 2032 - 1
=> x = 2031
đặt a=1/3+1/6+1/10+...........+2/n(n+1)
1/2a=1/6+1/12+...........+1/n(n+1)
1/2a=1/2.3+1/3.4+........+1/n(n+1)
1/2a=1/2-1/3+1/3-1/4+.......+1/n-1/n+1
1/2a=1/2-1/n+1
a=(1/2--1/n+1):1/2=2003/2004
1/2-1/n+1=2003/2004.1/2
1/2-1/n+1=2003/4008
1/n+1=1/2-2003/4008
1/n+1=1/4008
suy ra n+1=4008
n=4007
Bạn tham khảo câu trả lời tương tự ở đây nhé:
Câu hỏi của Nguyễn Hải - Toán lớp 7 - Học toán với OnlineMath
\(\frac{1}{3}\)+\(\frac{1}{6}\)+\(\frac{1}{10}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
\(\frac{2}{6}\)+\(\frac{2}{12}\)+\(\frac{2}{20}\)+...+\(\frac{2}{n\left(n+1\right)}\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{n.\left(n+1\right)}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{3}\)+\(\frac{1}{3}\)_\(\frac{1}{4}\)+\(\frac{1}{4}\)_\(\frac{1}{5}\)+...+\(\frac{1}{n}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
2\(\times\)\((\)\(\frac{1}{2}\)_\(\frac{1}{n+1}\)\()\)=\(\frac{2017}{2019}\)
\(\frac{1}{2}\)_\(\frac{1}{n+1}\)=\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2}\)_\(\frac{2017}{4038}\)
\(\frac{1}{n+1}\)=\(\frac{1}{2019}\)
\(\Rightarrow\)n+1=2019
\(\Rightarrow\)n=2018\(\in\)Z
Vậy n=2018
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)
\(\frac{1.2}{3.2}+\frac{1.2}{6.2}+\frac{1.2}{10.2}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)
\(\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{1}{n.\left(n+1\right)}=\frac{2015}{2016}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{1}{n\left(n+1\right)}=\frac{2015}{2016}\)
\(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{2015}{2016}\)
\(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)
\(2.\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{2015}{2016}\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{2016}:2\)
\(\frac{1}{2}-\frac{1}{n+1}=\frac{2015}{4032}\)
\(\frac{1}{n+1}=\frac{1}{2}-\frac{2015}{4032}\)
\(\frac{1}{n+1}=\frac{1}{4032}\)
\(\Rightarrow n+1=4032\)
\(\Rightarrow n=4031\)