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theo đề bài ta có
1+ 1/3 +1/6+...+2/x(x+1)=1+2009/2010
=>1/3+1/6+....+2/x(x+1)=2009/2010
=>1/2(2+1):2+1/3(3+1):2+.....+1/x(x+1):2=2009/2010
=>2/2(2+1)+2(3+1)+....+2/x(x+1)=2009/2010
=>2(1/2.3+1/3.4+....+1/x(x+1)=2009/2010
=>1/2-1/3+1/3-1/4+.....+1/x-1/x+1=2009/2010:2
=>1/2-1/x+1=2009/4020
=>1/x+1=1/2-2009/4020
=>1/x+1=1/4020
=>x+1=4020
=>x=4020-1
=>x=4019
`Answer:`
`1/3+1/6+1/10+...+2/(x.(x+1))=2008/2010`
`=2/6+2/12+2/20+...+2/(x.(x+1))=2008/2010`
`=2/(2.3)+2/(3.4)+2/(4.5)+...+(2)/(x.(x+1))=2008/2010`
`=2.(1/2-1/3+1/3-1/4+...+1/x(x+1))=2008/2010`
`=1/2-1/3+1/3-1/4+...+1/x-1/(x+1)=1004/2010`
`=1/2-1/(x+1)=1004/2010`
`=>1/(x+1)=1/2-1004/2010`
`=>1/(x+1)=1/2010`
`=>x+1=2010`
`=>x=2010-1`
`=>x=2009`
1/3+1/6+1/10+2/x.(x+1)=2010/1006
1/6+1/12+1/20+1/x.(x+1)=2010/2012
1/2.3+1/3.4+1/4.5+1/x.(x+1)=1005/1006
1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/x.(x+1)=1005/1006
1/2 - 1/5 + 1/x.(x+1)=1005/1006
3/10+1/x.(x+1)=1005/1006
1/x.(x+1)=1005/1006 - 3/10
1/x.(x+1)=1758/2515
x.(x+1)=1:1758/2515
x.(x+1)=2515/1758
Đến đây thì mình chịu òi!
Ta có : 1/3+1/6+1/10+ .....+2/x.(x+1)=2010/2012
=>2/6+2/12+2/20+........+2/x(x+1)=2010/2012
=>2.(1/2.3+1/3.4+1/4.5+.....+1/x.(x+1)=2010/2012
................................
Bạn tự làm tiếp nhé ! x=1005
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1\frac{2008}{2010}\):2
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}\)
\(\Rightarrow1-\frac{2009}{2010}=\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2010}=\frac{1}{x+1}\)
\(\Rightarrow x=2009\)
nha !
Ta có :A=1+\(\frac{2}{6}\)+\(\frac{2}{12}\)+......+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{4018}{2010}\)
\(\Rightarrow\)A=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)+...+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{x}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=\(\frac{1}{2}\)-\(\frac{1}{x+1}\)=\(\frac{502}{1005}\)
\(\Rightarrow\)\(\frac{1}{x+1}\)=\(\frac{1}{2010}\)\(\Rightarrow\)x+1=2010\(\Rightarrow\)x=2009
1.1/3+1/6+1/10+...+2/x.(x+1)=2007/2009
=>2/6+2/12+2/20+...+2/x.(x+1)=2007/2009
=>1/2-1/3+1/3-1/4+1/4-1/5+...+1/x-1/(x+1)=2007/2009:2
=>1/2-1/(x+1)=2007/4018
=>1/(x+1)=1/2-2007/4018
=>1/x+1=1/2009
=>x+1=2009
=>x=2009-2008
=>x=1
vậy x=1
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2010}\)
\(\Rightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2010}\)
\(\Rightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2010}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2010}\)
\(\Rightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2010}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{2010}:2\)
\(\Rightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2009}{4020}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4020}\)
\(\Rightarrow x+1=4020\)
=> x = 4020 - 1
=> x = 4019
1) \(A=\frac{7}{10\times11}+\frac{7}{11\times12}+\frac{7}{12\times13}+...+\frac{7}{69\times70}\)
\(A=7\times\left(\frac{1}{10\times11}+\frac{1}{11\times12}+\frac{1}{12\times13}+...+\frac{1}{69\times70}\right)\)
\(A=7\times\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{69}-\frac{1}{70}\right)\)
\(A=7\times\left(\frac{1}{10}-\frac{1}{70}\right)\)
\(A=7\times\frac{3}{35}\)
\(A=\frac{3}{5}\)
2) \(B=\frac{1}{25\times27}+\frac{1}{27\times29}+\frac{1}{29\times31}+...+\frac{1}{73\times75}\)
\(B=\frac{1}{2}\times\left(\frac{2}{25\times27}+\frac{2}{27\times29}+\frac{2}{29\times31}+...+\frac{2}{73\times75}\right)\).
\(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{27}+\frac{1}{27}-\frac{1}{29}+\frac{1}{29}-\frac{1}{31}+...+\frac{1}{73}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\times\left(\frac{1}{25}-\frac{1}{75}\right)\)
\(B=\frac{1}{2}\times\frac{2}{75}\)
\(B=\frac{1}{75}\)
3) \(C=\frac{4}{2\times4}+\frac{4}{4\times6}+\frac{4}{6\times8}+...+\frac{4}{2008\times2010}\)
\(C=\frac{4}{2}\times\left(\frac{2}{2\times4}+\frac{2}{4\times6}+\frac{2}{6\times8}+...+\frac{2}{2008\times2010}\right)\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{2008}-\frac{1}{2010}\right)\)
\(C=2\times\left(\frac{1}{2}-\frac{1}{2010}\right)\)
\(C=2\times\frac{502}{1005}\)
\(C=\frac{1004}{1005}\)
_Chúc bạn học tốt_
Hình như sai đề rồi.
1/1 > 2010/2011 rồi mà!
Nếu không sai đề thì không tìm được x