\(a,\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{43.45}=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{43.45}\right)=\frac{5}{3}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{43}-\frac{1}{45}\right)=\frac{5}{3}.\frac{44}{45}=\frac{44}{27}\)

\(\frac{5}{1\cdot3}+\frac{5}{3\cdot5}+...+\frac{5}{97\cdot99}=\frac{5}{2}\left[\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\right]\)

\(=\frac{5}{2}\left[\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right]=\frac{5}{2}\left[1-\frac{1}{99}\right]\)

\(=\frac{5}{2}\cdot\frac{98}{99}=\frac{245}{99}\)

\(=\frac{5}{2}\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{99}\right)\)

\(=\frac{5}{2}\times\frac{98}{99}\)

\(=\frac{245}{99}\)

=\(\frac{249}{99}\)

bn nhé

\(x+\left[\frac{5}{5}+\frac{5}{45}+\frac{5}{117}+\frac{5}{221}+...+\frac{5}{1845}\right]=2\)

\(x+\left[\frac{5}{1.5}+\frac{5}{5.9}+\frac{5}{9.13}+\frac{5}{13.17}+...+\frac{5}{41.45}\right]=2\)

\(\frac{5}{4}x+\frac{5}{4}.\left(\frac{4}{1.5}+\frac{4}{5.9}+\frac{4}{9.13}+\frac{4}{13.17}+...+\frac{4}{41.45}\right)=\frac{2.5}{4}=\frac{5}{2}\)

\(\frac{5}{4}x+\frac{5}{4}.\left(1-\frac{1}{45}\right)=\frac{5}{2}\)

\(\frac{5}{4}x+\frac{11}{9}=\frac{5}{2}\)

\(\frac{5}{4}x=\frac{23}{18}\)

\(x=\frac{46}{45}\)

\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}=5\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}\right)\)

\(=\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}\right)\)

\(=\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\right)\)

\(=\frac{5}{2}\left(1-\frac{1}{7}\right)=\frac{5}{2}\left(\frac{7}{7}-\frac{1}{7}\right)=\frac{5}{2}.\frac{6}{7}=\frac{15}{7}\)

a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

= \(1-\dfrac{1}{101}\)

=\(\dfrac{100}{101}\) 

\(\dfrac{5}{1.3}+\dfrac{5}{3.5}+\dfrac{5}{5.7}+...+\dfrac{5}{99.101}\)

=\(\dfrac{5}{2}.\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99+101}\right)\)

=\(\dfrac{5}{2}.\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\) 

=\(\dfrac{5}{2}.\left(1-\dfrac{1}{101}\right)\)

= \(\dfrac{5}{2}-\dfrac{100}{101}\)

= \(\dfrac{305}{202}\)

Bài 16: 

A = \(\dfrac{2n+1}{3n+1}\); đkxđ n \(\ne\) - \(\dfrac{1}{3}\)

Gọi ước chung lớn nhất của 2n + 1 là d

Ta có: 2n + 1 ⋮ d; 3n + 1 ⋮ d

2n + 1 ⋮ d ⇒ 3.(2n + 1) ⋮ d ⇒ 6n + 3 ⋮ d

3n + 1 ⋮ d ⇒ 2.( 3n+ 1) ⋮ d ⇒ 6n + 2 ⋮ d

 ⇒ 6n + 3 - (6n + 2) ⋮ d ⇒ 6n + 3 - 6n - 2⋮ d

                                     ⇒ 1 ⋮ d ⇒ d = 1

Ước chung lớn nhất của 2n + 1 và 3n + 1 là 1 

Hay 2n + 1 và 3n + 1 là hai số nguyên tố cùng nhau (đpcm)

M = 3/1x3 + 3/3x5 + 3/5x7 + ... + 3/45x47 + 3/47x49

M = 3/2 x (2/1x3 + 2/3x5 + 2/5x7 + ... + 2/45x47 + 2/47x49)

M = 3/2 x (1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/45 - 1/47 + 1/47 - 1/49)

M = 3/2 x (1 - 1/49)

M = 3/2 x 48/49

M = 72/49

N tính tương tự, nhân N với 5/4