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3 tháng 5 2018

\(\frac{1}{15}+\frac{1}{35}+\frac{1}{63}+...+\frac{1}{\left(2x+1\right)\left(2x+3\right)}=\frac{15}{93}\)

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

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

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

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

\(\frac{1}{3}-\frac{1}{2x+3}=\frac{15}{93}:\frac{1}{2}=\frac{10}{31}\)

\(\frac{1}{2x+3}=\frac{1}{3}-\frac{10}{31}=\frac{1}{93}\)

\(\Rightarrow2x+3=93\rightarrow2x=90\rightarrow x=45\)

6 tháng 5 2023

Bài giải chi tiết đây em nhé:

\(\dfrac{1}{3}\) + \(\dfrac{1}{15}\) + \(\dfrac{1}{35}\) + \(\dfrac{1}{63}\)+...+ \(\dfrac{1}{\left(2x-1\right)\left(2x+1\right)}\) = \(\dfrac{9}{19}\)

\(\dfrac{1}{2}\)(\(\dfrac{2}{1.3}\) + \(\dfrac{2}{3.5}\)+\(\dfrac{2}{5.7}\)\(\dfrac{2}{7.9}\)+...+ \(\dfrac{2}{\left(2x-1\right)\left(2x+1\right)}\)) = \(\dfrac{9}{19}\)

\(\dfrac{1}{2}\)\(\dfrac{1}{1}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\)\(\dfrac{1}{7}\) - \(\dfrac{1}{9}\) +... + \(\dfrac{1}{2x-1}-\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)

 \(\dfrac{1}{2}\) ( 1 - \(\dfrac{1}{2x+1}\)) = \(\dfrac{9}{19}\)

      1   - \(\dfrac{1}{2x+1}\) = \(\dfrac{9}{19}\) : \(\dfrac{1}{2}\)

       1  -  \(\dfrac{1}{2x+1}\) = \(\dfrac{18}{19}\)

               \(\dfrac{1}{2x+1}\) = \(1-\dfrac{18}{19}\)

                \(\dfrac{1}{2x+1}\) = \(\dfrac{1}{19}\)

                 \(2x+1\)  = 19

                 2\(x\)        = 19 - 1

                 2\(x\)       = 18

                    \(x\)      = 18: 2

                     \(x\)     = 9

 

8 tháng 8 2017

\(\dfrac{2x}{15}+\dfrac{2x}{35}+\dfrac{2x}{63}+...+\dfrac{2x}{195}=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+...+\dfrac{2}{195}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{13\cdot15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{13}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\left(\dfrac{1}{3}-\dfrac{1}{15}\right)=\dfrac{4}{5}\\ x\cdot\dfrac{4}{15}=\dfrac{4}{5}\\ x=\dfrac{4}{5}:\dfrac{4}{15}\\ x=3\)

Gọi \(D=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\)

\(2D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\\ 2D+D=\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\right)\\ 3D=1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}-\dfrac{1}{32}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}\\ 3D=1-\dfrac{1}{64}< 1\\ \Rightarrow D=\dfrac{1-\dfrac{1}{64}}{3}< \dfrac{1}{3}\)

Vậy \(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{8}-\dfrac{1}{16}+\dfrac{1}{32}-\dfrac{1}{64}< \dfrac{1}{3}\)

DD
26 tháng 11 2021

\(C=\frac{1}{3\times5}+\frac{1}{5\times7}+\frac{1}{7\times9}+...+\frac{1}{\left(2x+1\right)\times\left(2x+3\right)}\)

\(=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{\left(2x+1\right)\times\left(2x+3\right)}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2x+1}-\frac{1}{2x+3}\right)\)

\(=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{2x+3}\right)=\frac{15}{93}\)

\(\Leftrightarrow2x+3=93\)

\(\Leftrightarrow x=45\).

25 tháng 9 2018

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

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

\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{\left(2x+1\right)\left(2x+3\right)}=\frac{10}{31}\)

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

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

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

\(\Rightarrow2x+3=93\)

\(2x=90\)

\(x=\frac{90}{2}=45\)

Vậy \(x=45\)

Mình biết làm này nhưng sợ bạn ko hiểu

2              

28 tháng 6 2020

Em kiểm tra lại đề nhé!

14 tháng 4 2017

Ta có :

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

\(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+..............+\dfrac{2}{\left(2x+1\right)\left(2x+3\right)}=\dfrac{30}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+..............+\dfrac{1}{2x+1}-\dfrac{1}{2x+3}=\dfrac{30}{93}\)

\(\dfrac{1}{3}-\dfrac{1}{2x+3}=\dfrac{30}{93}\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{30}{93}=\dfrac{1}{2x+3}\)

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

\(\Rightarrow2x+3=93\)

\(2x=90\)

\(\Rightarrow x=45\)

Vậy \(x=45\) là giá trị cần tìm

~ Chúc bn học tốt ~