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__BQT Lớp 6/7 Hỏi Đáp__

\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.(x+1)}=\frac{2007}{2009}\)

=> \(2.\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)

=> \(2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2017}{2019}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2017}{2019}\)

=> \(2.\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2007}{2009}\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{2009}:2\)

=> \(\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)

=> \(\frac{1}{x+1}=\frac{1}{2}-\frac{2017}{4018}\)

=> \(\frac{1}{x+1}=\frac{1}{2019}\)

Vì 1 = 1

=> x + 1 = 2019

=> x       = 2019 - 1

=> x       = 2018

tra

r lời 

x=2018 

chúc bn 

hc tốt

\(\frac{x+10}{2008}+\frac{x+9}{2009}=\frac{x+8}{2010}+\frac{x+7}{2011}\)

\(\left(\frac{x+10}{2008}+1\right)+\left(\frac{x+9}{2009}+1\right)=\left(\frac{x+8}{2010}+1\right)+\left(\frac{x+7}{2011}+1\right)\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}=\frac{x+2018}{2010}+\frac{x+2018}{2011}\)

\(\frac{x+2018}{2008}+\frac{x+2018}{2009}-\frac{x+2018}{2010}-\frac{x+2018}{2011}=0\)

\(x+2018\cdot\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)=0\)

mà \(\left(\frac{1}{2008}+\frac{1}{2009}-\frac{1}{2010}-\frac{1}{2011}\right)\ne0\)

\(\Rightarrow x+2018=0\)

\(\Rightarrow x=-2018\)

Vậy,.............

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 \(x=-2018\)

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Rightarrow\frac{x+1}{2009}+1+\frac{x+2}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+1010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Rightarrow\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)=\left(x+2010\right)\left(\frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\right)\)

\(\Rightarrow x+2010=0\) vì \(0< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}< \frac{1}{2000}+\frac{1}{1999}+\frac{1}{1998}\)

\(\Rightarrow x=-2010\)

                                                            Bài giải

\(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

\(\Rightarrow\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)=\left(\frac{x+10}{2000}+1\right)+\left(\frac{x+11}{1999}+1\right)+\left(\frac{x+12}{1998}+1\right)\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-(\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998})=0\)

\(\Rightarrow\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

\(\left(x+2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

Vì \(\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)\ne0\) nên \(x+2010=0\)

                                                                                                                          \(x=0-2010=-2010\)

\(\dfrac{x+1}{2009}+\dfrac{x+2}{2008}+\dfrac{x+3}{2007}=\dfrac{x+10}{2000}+\dfrac{x+11}{1999}+\dfrac{x+12}{1998}\)

\(\Rightarrow\left(\dfrac{x+1}{2009}+1\right)+\left(\dfrac{x+2}{2008}+1\right)+\left(\dfrac{x+3}{2007}+1\right)=\left(\dfrac{x+10}{2000}+1\right)+\left(\dfrac{x+11}{1999}+1\right)+\left(\dfrac{x+12}{1998}+1\right)\)

\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}=\dfrac{x+2010}{2000}+\dfrac{x+2010}{1999}+\dfrac{x+2010}{1998}\)\(\Rightarrow\dfrac{x+2010}{2009}+\dfrac{x+2010}{2008}+\dfrac{x+2010}{2007}-\dfrac{x+2010}{2000}-\dfrac{x+2010}{1999}-\dfrac{x+2010}{1998}=0\)\(\Rightarrow\left(x+2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2010}+\dfrac{1}{2007}-\dfrac{1}{2000}-\dfrac{1}{1999}-\dfrac{1}{1998}\right)=0\)\(\Rightarrow x+2010=0\Rightarrow x=-2010\)

bạn 3 năm nữa mới có người giải

đã hơn 3 năm rồi nhưng chưa có ai giải, mà 3 năm rồi bn cx ko cần nx.