3A=3/2.5+...+3/2018.2021

3A=1/2-1/5+1/5-...+1/2018-1/2021

3A=1/2-1/2021 sau tự tính A

3A= 1/2- 1/5 + 1/5- 1/8+ 1/8 -1/11+...+ 1/2012- 1/2015 +1/2015-  1/2018-1/2021

 3A   =1/2 -1/2021 

3A    = 2019/ 4042

  => 2019/4042 : 3 = 673/4042      

Chúc bạn học tốt !!

Giải:

A=1/2.5+1/5.8+1/8.11+...+1/2012.2015+1/2015.2018+1/2018.2021

A=1/3.(3/2.5+3/5.8+3/8.11+...+3/2012.2015+3/2015.2018+3/2018.2021)

A=1/3.(1/2-1/5+1/5+1/8+1/8-1/11+...+1/2012-1/2015+1/2015-1/2018+1/2018-1/2021)

A=1/3.(1/2-1/2021)

A=1/3.2019/4042

A=673/4042

Chúc bạn học tốt!

nhân 3 vào cả hai vế 

\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)

\(A=\dfrac{1}{2}-\dfrac{1}{98}=\dfrac{49}{98}-\dfrac{1}{98}=\dfrac{48}{98}=\dfrac{24}{49}\)

\(A=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\\ A=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)=\dfrac{1}{3}\cdot\dfrac{24}{49}=\dfrac{8}{49}\)

Đặt A=\(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\)

\(3A=\dfrac{3}{2.5}+\dfrac{3}{5.8}+...+\dfrac{3}{95.98}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\)

\(3A=\dfrac{1}{2}-\dfrac{1}{98}\)

\(3A=\dfrac{24}{49}\Rightarrow A=\dfrac{8}{49}\)

    \(\dfrac{1}{2.5}+\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{92.95}+\dfrac{1}{95.98}\)

\(=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\)

\(=\dfrac{1}{2}-\dfrac{1}{98}\)

\(=\dfrac{24}{49}\)

\(\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{92\cdot95}+\dfrac{3}{95\cdot98}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\)

\(=\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)

\(=\dfrac{1}{3}\cdot\dfrac{96}{196}=\dfrac{8}{49}\)

\(A=\dfrac{1}{2\cdot5}+\dfrac{1}{5\cdot8}+\dfrac{1}{8\cdot11}+...+\dfrac{1}{92\cdot95}+\dfrac{1}{95\cdot98}\)

\(A=\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{2}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{92}+\dfrac{1}{92}-\dfrac{1}{95}+\dfrac{1}{95}-\dfrac{1}{98}\)

\(A=\dfrac{1}{2}-\dfrac{1}{98}\)

\(A=\dfrac{49}{98}-\dfrac{1}{98}\)

\(A=\dfrac{48}{98}\)

\(A=\dfrac{24}{49}\)

Giải thích các bước giải:

A =1/2.5 + 1/5.8 + 1/8.11 + … +1/92.95 + 1/95.98

=1/3 . (1/2-1/5+1/5-1/8+1/8-1/11+…+1/92-1/95+1/95-1/98)

=1/3 . (1/2 – 1/98 )

=1/3 . 24/49

=8/49`

vậy `A=8/49`

\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{y\left(y+3\right)}=\dfrac{98}{1545}\)

\(\Leftrightarrow\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{y\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{y}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{y+3}{5\left(y+3\right)}-\dfrac{5}{5\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{y+3-5}{5\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{y-2}{5\left(y+3\right)}=\dfrac{98}{515}\)

\(\Leftrightarrow515\left(y-2\right)=98.5\left(y+3\right)\)

\(\Leftrightarrow515y-1030=490y+1470\)

\(\Leftrightarrow25y-2500=0\\ \Leftrightarrow25y=2500\\ \Leftrightarrow y=100\)

\(\Leftrightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{y+3}\right)=\dfrac{98}{1545}\)

\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{y+3}=\dfrac{98}{515}\)

\(\Leftrightarrow\dfrac{1}{y+3}=\dfrac{1}{103}\)

hay x=100

đề thiếu nhé

Ta có: \(\dfrac{k}{x.\left(x+k\right)}=\dfrac{x+k-x}{x.\left(x+k\right)}=\dfrac{1}{x}-\dfrac{1}{x+k}\)

nên áp dụng ta có:

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

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

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

Nên $\dfrac{1}{3}.\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{x.\left(x+3\right)}\right)=\dfrac{1}{3}.(\dfrac{1}{5}-\dfrac{1}{x+3})$
Đến đây là làm được rồi nha

\(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot17}\)

= \(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{17}\)

\(=\dfrac{1}{2}-\dfrac{1}{17}\)

\(=\dfrac{15}{34}\)

Vì \(\dfrac{15}{34}< \dfrac{1}{2}=>\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+\dfrac{3}{8\cdot11}+\dfrac{3}{11\cdot14}+\dfrac{3}{14\cdot27}< \dfrac{1}{2}\)

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

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

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

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

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

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

\(\Rightarrow x+3=308\)

\(\Rightarrow x=305\)

vậy \(x=305\)