đề bài đúng đấy bạn,hông sai đâu,có cần mk giải lun cho hông???
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+...+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{3}\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{x}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{x+3-5}{5\left(x+3\right)}=\dfrac{303}{1540}\)
\(\Rightarrow\dfrac{x-2}{5\left(x+3\right)}=\dfrac{303}{1540}\)
\(\Rightarrow1540\left(x-2\right)=303.5\left(x+3\right)\)
\(\Rightarrow1540x-3080=1515x+4545\)
\(\Rightarrow25x=7625\)
\(\Rightarrow x=305\)
Vậy x = 305
\(\dfrac{1}{5.8}+\dfrac{1}{8.11}+.....+\dfrac{1}{x\left(x+3\right)}=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{3}{5.8}+\dfrac{3}{8.11}+.....+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+....+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{3}\left(\dfrac{1}{5}-\dfrac{1}{x+3}\right)=\dfrac{101}{1540}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{101}{1540}:\dfrac{1}{3}\\ \Rightarrow\dfrac{1}{5}-\dfrac{1}{x+3}=\dfrac{303}{1540}\\ \Rightarrow\dfrac{1}{x+3}=\dfrac{1}{5}-\dfrac{303}{1540}\\ \Rightarrow\dfrac{1}{x+3}=\dfrac{1}{308}\\ \Rightarrow x+3=308\\ \Rightarrow x+=308-3=5\\ \\ \)