1 phần 1 x 3 + 1 phần 3 x 5+.......+1 phần 2009 x 2011
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S
16 tháng 4 2017
b, \(\frac{x+1}{2009}+\frac{x+2}{2009}=\frac{x+10}{2000}+\frac{x+11}{1999}\)
\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)\)
\(\Rightarrow\frac{x+1+2009}{2009}+\frac{x+2+2008}{2008}=\frac{x+10+2000}{2000}+\frac{x+11+1999}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}=\frac{x+2010}{2000}+\frac{x+2010}{1999}\)
\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}-\frac{x+2010}{2000}-\frac{x+2010}{1999}=0\)
\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\right)=0\)
Mà \(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2000}-\frac{1}{1999}\ne0\)
=> x + 2010 = 0 => x = -2010
15 tháng 4 2017
Đây là cuộc thi nhé. cần sự công bằng. Mong em không tái phạm lần sau. Bạn sẽ bị khóa nick hoặc trừ 5000 điểm nhé!
BQT thân gửi em!
__BQT Lớp 6/7 Hỏi Đáp__
J
3
XO
7 tháng 6 2019
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.(x+1)}=\frac{2007}{2009}\)
=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)
=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2\)
=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4018}\)
=> \(\frac{1}{x+1}=\frac{1}{2019}\)
Vì 1 = 1
=> x + 1 = 2019
=> x = 2019 - 1
=> x = 2018
PN
6 tháng 8 2020
\(\frac{5}{6}=\frac{x-1}{x}\left(đk:x\ne0\right)\)
\(< =>5x=6\left(x-1\right)< =>5x=6x-6\)
\(< =>6x-5x=6< =>x=6\left(tmđk\right)\)
\(\frac{1}{2}=\frac{x+1}{3x}\left(đk:x\ne0\right)\)
\(< =>3x=2\left(x+1\right)< =>3x=2x+2\)
\(< =>3x-2x=2< =>x=2\left(tmđk\right)\)
\(\frac{3}{x+2}=\frac{5}{2x+1}\left(đk:x\ne-2;-\frac{1}{2}\right)\)
\(< =>3\left(2x+1\right)=5\left(x+2\right)< =>6x+3=5x+10\)
\(< =>6x-5x=10-3< =>x=7\left(tmđk\right)\)
\(\frac{5}{8x-2}=-\frac{4}{7-x}\left(đk:x\ne\frac{1}{4};7\right)\)
\(< =>\frac{5}{8x-2}=\frac{4}{x-7}< =>5\left(x-7\right)=4\left(8x-2\right)\)
\(< =>5x-35=32x-8< =>32x-5x=-35+8\)
\(< =>27x=-27< =>x=-1\)
PN
6 tháng 8 2020
\(\frac{4}{3}=\frac{2x-1}{3}< =>4.3=\left(2x-1\right).3\)
\(< =>12=6x-3< =>6x=12+3\)
\(< =>6x=15< =>x=\frac{15}{6}=\frac{5}{2}\)
\(\frac{2x-1}{3}=\frac{3x+1}{4}< =>4\left(2x-1\right)=3\left(3x+1\right)\)
\(< =>8x-4=9x+3< =>9x-8x=-4-3\)
\(< =>9x-8x=-7< =>x=-7\)
\(\frac{4}{x+2}=\frac{7}{3x+1}\left(đk:x\ne-2;-\frac{1}{3}\right)\)
\(< =>4\left(3x+1\right)=7\left(x+2\right)< =>12x+4=7x+14\)
\(< =>12x-7x=14-4< =>5x=10\)
\(< =>x=\frac{10}{5}=2\left(tmđk\right)\)
\(-\frac{3}{x+1}=\frac{4}{2-2x}\left(đk:x\ne-1;1\right)\)
\(< =>-3\left(2-2x\right)=4\left(x+1\right)< =>-6+6x=4x+4\)
\(< =>6x-4x=4+6< =>2x=10\)
\(< =>x=\frac{10}{2}=5\left(tmđk\right)\)
\(\frac{x+1}{3}=\frac{3}{x+1}\left(đk:x\ne-1\right)\)
\(< =>\left(x+1\right)\left(x+1\right)=3.3\)
\(< =>x^2+2x+1=9< =>x^2+2x+1-9=0\)
\(< =>x^2+2x-8=0< =>x^2-2x+4x-8=0\)
\(< =>x\left(x-2\right)+4\left(x-2\right)=0< =>\left(x+4\right)\left(x-2\right)=0\)
\(< =>\orbr{\begin{cases}x+4=0\\x-2=0\end{cases}< =>\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\left(tmđk\right)\)
LV
3
3 tháng 9 2018
\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\left(\frac{x+10}{2008}+1\right)+\left(\frac{x+9}{2009}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)
\(\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
mà \(\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)\ne0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
Vậy,.............
D
3 tháng 9 2018
Ta có: \(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)
\(\Rightarrow\frac{x+10}{2008}+1+\frac{x+9}{2009}+1=\frac{x+8}{2010}+1+\frac{x+7}{2011}+1\)
\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)
\(\Rightarrow\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)
\(\Rightarrow x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)
Do \(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\ne0\)
\(\Rightarrow x+2018=0\)
\(\Rightarrow x=-2018\)
Vậy \(x=-2018\)
ND
0
gọi A=\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{2009.2011}\)
A=\(\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{2009}-\frac{1}{2011}\right)\)
A=\(\frac{1}{2}.\left(1-\frac{1}{2011}\right)\)
A=\(\frac{1005}{2011}\)
bài có sai sót gì thì bảo tớ nhé!