bn lấy 1/2 nhân ra ngoài ròi tính như bình thường nha!

Đặt tổng trên là A ta có

\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)

\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)

\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)

\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
 

\(\frac{1}{10.12}+\frac{1}{12.14}+\frac{1}{14.16}+...+\frac{1}{48.50}\)

=\(\frac{1}{5.2.2.6}+\frac{1}{6.2.2.7}+\frac{1}{7.2.2.8}+...+\frac{1}{24.2.2.25}\)

=\(\frac{1}{2}.\left(\frac{2}{5.6}+\frac{2}{6.7}+\frac{2}{7.8}+...+\frac{2}{24.25}\right)\)

=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{24}-\frac{1}{25}\right)\)

=\(\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{25}\right)\)

=\(\frac{1}{2}.\frac{4}{25}\)

=\(\frac{2}{25}\)

mình không biết đúng hông có gì sai cho mình xin lỗi

=3/2(2/10.12+2/12.14+...+2/48.50)

=3/2(1/10-1/12+1/12-1/14+...+1/48-1/50)

=3/2(1/10-1/50)

=3/2 . 2/25 =3/25

Đặt phép tính trên là A

Ta có:

A=3/10*12+3/12*14+3/14*16+...+3/48*50

A*2/3=2/10*12+2/12*14+2/14*16+...+2/48*50

A*2/3=1/10-1/12+1/12-1/14+1/14-1/16+...+1/48-1/50

A*2/3=1/10-1/50

A*2/3=2/25

A=2/25:2/3

A=3/25

Vậy A=3/25

Nếu đúng thì k cho mình nha

\(\frac{3}{10.12}+\frac{3}{12.14}+\frac{3}{14.16}+...+\frac{3}{48.50}\)

\(=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{50}\right)\)

\(=\frac{3}{2}.\frac{2}{25}\)

\(=\frac{3}{25}\)

S=\(\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+.....+\frac{2}{98.100}\)

S=\(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+........+\frac{1}{98}-\frac{1}{100}\)

S=\(\frac{1}{10}-\frac{1}{100}\)

S=\(\frac{9}{100}\)<\(\frac{1}{10}\)

tính S = cánh tính sai phân  

con cac

\(D=\frac{2}{25.27}+2\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(D=2.\left(\frac{1}{25}-\frac{1}{27}\right)+2\left(\frac{1}{10}-\frac{1}{50}\right)\)

\(D=2.\frac{2}{675}+2.\frac{2}{25}\)

\(D=2.\left(\frac{2}{675}+\frac{2}{25}\right)\)

\(D=2.\frac{56}{675}\)

\(D=\frac{112}{675}\)

Study well 

Đặt \(A=\frac{3}{10.12}+\frac{3}{12.14}+.....+\frac{3}{48.50}\)

      \(A=\frac{3}{2}.\left(\frac{2}{10.12}+\frac{2}{12.14}+......+\frac{2}{48.50}\right)\)

      \(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{12}+....+\frac{1}{48}-\frac{1}{50}\right)\)

       \(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{50}\right)\)

      \(A=\frac{3}{2}.\frac{2}{25}\) 

     \(A=\frac{3}{25}\)

cảm ơn bạn nhiều nhé !
 

   3/10.12+3/12.14+............+3/48.50

=3/2.(2/10.12+2/12.14+..........+2/48.50)

=3/2(2/10-2/12+2/12-2/14+......+2/48-2/50)

=3/2.(2/10-2/50)

=3/2.4/25

=6/25

Ta có: A=\(\frac{1}{10\cdot12}+\frac{1}{12\cdot14}+\frac{1}{14\cdot16}+...+\frac{1}{38\cdot40}\)

=> \(A=\frac{1}{4}\cdot\left(\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+...+\frac{1}{19\cdot20}\right)\)

          =\(\frac{1}{4}\cdot\left(\frac{6-5}{5\cdot6}+\frac{7-6}{6\cdot7}+\frac{8-7}{7\cdot8}+...+\frac{20-19}{19\cdot20}\right)\)

         =  \(\frac{1}{4}\cdot\left(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{19}-\frac{1}{20}\right)\)

         = \(\frac{1}{4}\cdot\left(\frac{1}{5}-\frac{1}{20}\right)\)

         = \(\frac{1}{4}\cdot\frac{3}{20}=\frac{3}{80}\)

Vậy A= 3/80

A = 1/10 - 1/12 + 1/12 - 1/14 + ....+ 1/38 - 1/40
A = 1/10 - 1/40
A = 4/40 - 1/40
A = 3/40
Chúc bạn học tốt !

(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì 2n⋮n-2

2n-4+4⋮n-2

2n-4⋮n-2⇒4⋮n-2

n-2∈Ư(4)⇒Ư(4)={1;-1;2;-2;4;-4}

n∈{3;1;4;0;6;-2}

(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+...+\dfrac{3}{48.50}\)

=\(\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+...+\dfrac{2}{48.50}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

=\(\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\)

=\(\dfrac{3}{2}.\dfrac{2}{25}\)

=\(\dfrac{3}{25}\)

Giải:

(1) Để \(\dfrac{2n}{n-2}\) là số nguyên thì \(2n⋮n-2\) 

\(2n⋮n-2\) 

\(\Rightarrow2n-4+4⋮n-2\) 

\(\Rightarrow4⋮n-2\) 

\(\Rightarrow n-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\) 

Vậy \(n\in\left\{0;1;3;4;6\right\}\)

(2) \(\dfrac{3}{10.12}+\dfrac{3}{12.14}+\dfrac{3}{14.16}+...+\dfrac{3}{48.50}\) 

\(=\dfrac{3}{2}.\left(\dfrac{2}{10.12}+\dfrac{2}{12.14}+\dfrac{2}{14.16}+...+\dfrac{2}{48.50}\right)\) 

\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\) 

\(=\dfrac{3}{2}.\left(\dfrac{1}{10}-\dfrac{1}{50}\right)\) 

\(=\dfrac{3}{2}.\dfrac{2}{25}\) 

\(=\dfrac{3}{25}\) 

Chúc bạn học tốt!

(1) Để biểu thức \(\dfrac{2n}{n-2}\) nguyên thì \(2n⋮n-2\)

\(\Leftrightarrow4⋮n-2\)

\(\Leftrightarrow n-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(n\in\left\{3;1;4;0;6;-2\right\}\)

\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)

\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)

\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)

\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)

\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)

\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)

\(=\frac{2}{5}.\frac{12}{25}\)

\(=\frac{24}{125}\)

S=1/5.6+1/10.9+1/15.12+...+1/3350.2013

 =(1/5).(1/3).(1/1.2+1/2.3+1/3.4+...+1/670.671)

 =(1/15). (1-1/2+1/2-1/3+...+1/670-1/671)

 =(1/15). (1-1/671)

 =1/15.670/671

 =134/2013