D=1/1.5+1/5.9+...+1/41.45

4D=4/1.5+4/5.9+...+4/41.45

4D=1-1/5+1/5-1/9+...+1/41-1/45

4D=1-1/45

D=44/45:4=11/45

\(\text{Đề bài sai : }\frac{4}{\left(n-4\right)^n}->\frac{4}{\left(n-4\right)^n}\)

\(\text{Ta có :}\)

                                               \(S=\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-...-\frac{4}{\left(n-4\right)n}\)

                                                  \(=\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)

                                                  \(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-...-\frac{1}{n-4}+\frac{1}{n}\)

                                                  \(=\frac{1}{1}-\frac{1}{5}-\frac{1}{5}+\frac{1}{n}\)

                                                  \(=\frac{3}{5}+\frac{1}{n}\)

                                                  \(=\frac{3}{5}+\frac{1}{n}\)

                                                  \(=\frac{3n+5}{5n}\)

\(\text{Vậy ...}\)

\(a,x=\dfrac{13}{2}-2\\ x=\dfrac{9}{2}\\ b,x=\dfrac{4}{5}\times\dfrac{3}{4}\\ x=\dfrac{12}{20}=\dfrac{3}{5}\)

Có dạng tổng quát như thế này nhé: 
\(\frac{k}{n\left(n+k\right)}=\frac{1}{n}-\frac{1}{k+n}\)

Trong trường hợp này là \(\frac{-4}{1.5}-\frac{4}{5.9}-...-\frac{4}{\left(n+4\right)n}=-\left(\frac{1}{1}-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{n}-\frac{1}{n+4}\right)\)

Đáp án là: \(\frac{1}{n+4}-1\)

\(\dfrac{2021}{1\cdot5}+\dfrac{2021}{5\cdot9}+...+\dfrac{2021}{x\cdot\left(x+4\right)}=505\)

\(2021\cdot\left(\dfrac{1}{1.5}+\dfrac{1}{5\cdot9}+...+\dfrac{1}{x\cdot\left(x+4\right)}\right)=505\)

\(\dfrac{2021}{4}\cdot\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{x\cdot\left(x+4\right)}\right)=505\)

\(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)

\(1-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)

\(\dfrac{1}{x+4}=\dfrac{1}{2021}\)

=> \(x+4=2021\)

=> \(x=2017\)

vậy \(x=2017\)

Ta có: \(\dfrac{2021}{1\cdot5}+\dfrac{2021}{5\cdot9}+...+\dfrac{2021}{x\left(x+4\right)}=505\)

\(\Leftrightarrow\dfrac{2021}{4}\left(\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{x\left(x+4\right)}\right)=505\)

\(\Leftrightarrow1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{x}-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)

\(\Leftrightarrow-\dfrac{1}{x+4}=\dfrac{2020}{2021}\)

\(\Leftrightarrow x+4=\dfrac{-2021}{2020}\)

hay \(x=-\dfrac{10101}{2020}\)