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: \(\frac{1}{n+4}-1\)

M = - ( 4/1.5 + 4/5.9 + ..................+ 4/(n-4).n

M = - (1-1/5 + 1/5 - 1/9 +..............+1/(n-4) - 1/n

M = -(1-1/n)

M = -1 + 1/n 

M = -n + 1

xie xie bn nha

Ta có : \(-\frac{4}{1.5}-\frac{4}{5.9}-\frac{4}{9.13}-.....-\frac{4}{\left(n+4\right)n}\)

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

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

\(=-\left(1-\frac{1}{n+4}\right)\)

\(=-\left(\frac{n+4}{n+4}-\frac{1}{n+4}\right)\)

\(=-\frac{n+3}{n+4}\)

A=1-1/5+1/5-1/9+...+1/(n-4)-1/n

A=1-1/n

A=n-1/n

= 1-1/5+1/5-1/9+1/9-1/13+...+1/n-4-1/n

=1-1/n

= n-1/n

đồ nguyễn việt anh bắt chước

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

\(=-\left(\frac{1}{1}-\frac{1}{5}\right)-\left(\frac{1}{5}-\frac{1}{9}\right)-\left(\frac{1}{9}-\frac{1}{13}\right)-...-\left(\frac{1}{n-4}-\frac{1}{n}\right)\)

\(=-\frac{1}{1}+\frac{1}{5}-\frac{1}{5}+\frac{1}{9}-\frac{1}{9}+\frac{1}{13}-...-\frac{1}{n-4}+\frac{1}{n}\)

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

\(-\left(\dfrac{4}{1.5}+\dfrac{4}{5.9}+\dfrac{4}{9.13}...+\dfrac{4}{n\left(n+4\right)}\right)\) \(=-\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{n}+\dfrac{1}{n+4}\right)=-\left(1-\dfrac{1}{n+4}\right)=-1+\dfrac{1}{n+4}\)