1-1/6+1/6-1/11+....+1/(5x+1)-1/(5x+2)=2010/2011                                                                                       <=>1-1/(5x+2)=2010/2011                                                                                                                               <=>1/2011=1/(5x+2)                                                                                                                                       <=>x=401

=2/3.5+2/5.7+2/7.9+........+2/(2x+1).(2x+3)=15/93.2

=1/3-1/5+1/5-1/7+.........+1/(2x+1)-1/(2x+3)=30/93

=1/3-1/(2x+3)=30/93

=1/(2x+3)=1/3-30/93

1/(2x+3)=31/93-30/93

1/(2x+3)=1/93

(2x+3).1=1.93

2x+3=93

2x=93-3

2x=90

x=90:2

x=45

uuwugtwf

2/3.5+2/5.7+2/7.9+...+2/(2x+1)(2x+3)=2.15/93

1/3-1/5+1/5-1/7+...+1/2x+1-1/2x+3=10/31

1/3-1/2x+3=10/31

1/(2x+3)=1/93

2x+3=93

2x=90

x=45

Mình biết làm này nhưng sợ bạn ko hiểu

2              

nhớ tk cho mk bài này như sau:

  1/2.[2/3.5+2/5.7+2/7.9+...+2/(2x+1).(2x+3)]=15/93

  1/2.[1.3-1/5+1/5-1/7+1/7-1/9+...+1/(2x+1)-1/(2x+3)]=5/31(rút gọn)

  1/3-1/(2x+3)=5/31:1/2

  1/3-1/(2x+3)=10/31

  1/2x+3=1/3-10/31

  1/2x+3=1/93

=>2x+3=93

    2x=93-3

    2x=90

      x=90:2

      x=40

 Vậy x=40

Ta có \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)(đk : \(x\ne0\))

=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)

=> \(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)

=> \(\frac{7}{x}=\frac{7}{15}\)

=> x = 15 (tm)

b) \(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

=> \(\frac{1}{2}\left(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}\right)=\frac{15}{93}\)

=> \(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}=\frac{10}{31}\)

=> \(\frac{1}{3}-\frac{1}{n+3}=\frac{10}{31}\)

=> \(\frac{1}{2x+3}=\frac{1}{93}\)

=> 2x + 3 = 93

=> 2x = 90

=> x = 45 

Ta có:

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}\)

\(=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}\)

\(=1-\frac{1}{5x+6}\)

\(=\frac{5x+5}{5x+6}=\frac{2010}{2011}\)

\(\Rightarrow5x+5=2010\)

\(\Rightarrow5x=2010-5=2005\)

\(\Rightarrow x=2005:5=401\)

Vậy x=401

=> 1 - 1/6 + 1/6 - 1/11 + ......+ 1/5x+1 - 1/5x + 6 = 2010/2011

=> 1 - 1/5x+6 = 2010/2011

=> 1/5x+6 = 1 - 2010/2011

=> 1/5x + 6 = 2009/2011

=> ...........................Còn lại bạn tự làm nha!

Ai k mk mk k lại !!!

\(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right).\left(5x+6\right)}=\frac{2010}{2011}\)

\(\Rightarrow1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\Rightarrow1-\frac{1}{5x+6}=\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{5x+6}=1-\frac{2010}{2011}\)

\(\Rightarrow\frac{1}{5x+6}=\frac{1}{2011}\)

\(\Rightarrow5x+6=2011\)

\(\Rightarrow5x=2011-6\)

\(\Rightarrow5x=2005\)

\(\Rightarrow x=401\)

b,\(\dfrac{1}{3.5}+\dfrac{1}{5.7}\)\(+\dfrac{1}{7.9}+....+\dfrac{1}{\left(2x+1\right).\left(2x+3\right)}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left[\dfrac{1}{3}+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+\left(\dfrac{1}{7}-\dfrac{1}{7}\right)+....+\left(\dfrac{1}{2x+1}-\dfrac{1}{2x+1}\right)-\dfrac{1}{2x+3}\right].\dfrac{1}{2}=\dfrac{15}{93}\)

\(\left(\dfrac{1}{3}+0+0+...+0-\dfrac{1}{2x+3}\right).\dfrac{1}{2}=\dfrac{15}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{15}{93}:\dfrac{1}{2}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{3}-\dfrac{10}{31}\)

\(\dfrac{1}{2x+3}=\dfrac{1}{93}\)

\(\Rightarrow2x+3=93\)

\(2x=93-3=90\)

\(\Rightarrow x=90:2=45\)