.....

<=>\(\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}\right)=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

<=>\(\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

<=>\(\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)

<=>\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

<=> x+3=308

<=> x=305

Giải:

a)1/5.8+1/8.11+...+1/x.(x+1)=101/1540

1/3.(3/5.8+3/8.11+...+3/x.(x+1))=101/1540

1/3.(1/5-1/8+1/8-1/11+...+1/x-1/x+1)=101/1540

                              1/3.(1/5-1/x+1)=101/1540

                                      1/5-1/x+1=101/1540:1/3

                                      1/5-1/x+1=303/1540

                                            1/x+1=1/5-303/1540

                                            1/x+1=1/308

⇒x+1=308

       x=308-1

       x=307

b)1/1.2+1/2.3+1/3.4+...+1/x.(x+1)=2020/2021

1/1-1/2+1/2-1/3+1/3-1/4+...+1/x-1/x+1=2020/2021

                                    1/1-1/x+1=2020/2021

                                          1/x+1=1/1-2020/2021

                                          1/x+1=1/2021

⇒x+1=2021

      x=2021-1

      x=2020

Mk thấy đề bài hơi sai là:

1/x+(x+1) ➜ 1/x.(x+1)

mới ra đc kết quả!

\(\frac{2}{40}+\frac{2}{88}+\frac{2}{154}+..+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{5.8}+\frac{2}{8.11}+\frac{2}{11.14}+...+\frac{2}{x\left(x+3\right)}=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

\(\Leftrightarrow\frac{2}{3}\left(\frac{1}{5}-\frac{1}{x+3}\right)=\frac{202}{1540}\)

\(\Leftrightarrow\frac{1}{5}-\frac{1}{x+3}=\frac{202}{1540}:\frac{2}{3}=\frac{303}{1540}\)

\(\Leftrightarrow\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}=\frac{1}{308}\)

\(\Rightarrow x+3=308\Rightarrow x=305\)

Vạy x = 305

a/ \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^{10}}\)

\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.......+\dfrac{1}{2^9}\)

\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+......+\dfrac{1}{2^9}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+.....+\dfrac{1}{2^{10}}\right)\)

\(\Leftrightarrow A=1-\dfrac{1}{2^{10}}\)

b/ \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+.......+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

\(\Leftrightarrow3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+......+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{101}{1540}.3\)

\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+......+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+.....+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\Leftrightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\)

\(\Leftrightarrow x+3=308\)

\(\Leftrightarrow x=305\)

Vậy ..

c/ \(1+\dfrac{1}{3}+\dfrac{1}{6}+........+\dfrac{1}{x\left(x+1\right):2}=1\dfrac{2007}{2009}\)

\(\dfrac{1}{2}\left(\dfrac{1}{3}+\dfrac{1}{6}+.......+\dfrac{1}{x\left(x+1\right):2}\right)=\dfrac{4016}{2009}.\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(\Leftrightarrow\dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{x\left(x+1\right)}=\dfrac{2008}{2009}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{2008}{2009}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{2009}\)

\(\Leftrightarrow x+1=2009\)

\(\Leftrightarrow x=2008\)

Vậy ..

bài 1:

A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{10}}\)

ta thấy 2A=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^9}\)

=>2A-A=\(1-\dfrac{1}{2^{10}}=\dfrac{1023}{1024}\)

a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)

                \(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)

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

                       \(x=\dfrac{3}{2}-\dfrac{3}{2}\)

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

a)

<=> (1/3)[3/(5.8) + 3/(8.11) + ... + 3/[x(x+3)] = 101/1540
<=> (1/3)[(1/5 - 1/8) + (1/8 - 1/11) + ... + 1/x - 1/(x+3)] = 101/1540
<=> (1/3)[1/5 - 1/(x+3)] = 101/1540
<=> 1/5 - 1/(x+3) = 303/1540
<=> 1/(x+3) = 1/5 - 303/1540 = 5/1540 = 1/308
<=> x = 305

b)

a)\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x.\left(x+3\right)}=\dfrac{101}{1540}\)

\(\dfrac{1.3}{5.8}+\dfrac{1.3}{8.11}+\dfrac{1.3}{11.14}+...+\dfrac{1.3}{x.\left(x+3\right)}=\dfrac{101.3}{1540}\)

\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}=\dfrac{303}{1540}\)

\(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)

\(\dfrac{1}{x+3}=\dfrac{1}{308}\)

308.1 = (x + 3).1

308 = x + 3

x = 308 - 3

x = 305

a) \(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)

=> \(3.\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{x\left(x+3\right)}\right)=\dfrac{101}{1540}.3\)

=> \(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{303}{1540}\)

=> \(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

=> \(\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)

=> \(\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\)

=> \(\dfrac{1}{x+3}=\dfrac{1}{308}\)

=> \(x+3=308\)

=> \(x=305\)

\(\dfrac{1}{x^2+7x+10}+\dfrac{1}{x^2+13x+40}+\dfrac{1}{x^2+19x+88}+\dfrac{1}{x^2+25x+154}\)

\(=\dfrac{1}{\left(x+2\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+8\right)}+\dfrac{1}{\left(x+8\right)\left(x+11\right)}+\dfrac{1}{\left(x+11\right)\left(x+14\right)}\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+11}+\dfrac{1}{x+11}-\dfrac{1}{x+14}\)

\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}\)