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Đặt tổng trên là A ta có
\(2A=\frac{2}{10.12}+\frac{2}{12.14}+\frac{2}{14.16}+...+\frac{2}{48.52}\)
\(2A=\frac{12-10}{10.12}+\frac{14-12}{12.14}+\frac{16-14}{14.16}+...+\frac{50-48}{48.50}\)
\(2A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+\frac{1}{14}-\frac{1}{16}+...+\frac{1}{48}-\frac{1}{50}=\frac{1}{10}-\frac{1}{50}=\frac{2}{25}\)
\(\Rightarrow A=\frac{2A}{2}=\frac{1}{25}\)
=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
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=\(\frac{9}{100}\)<\(\frac{1}{10}\)
Đặt \(A=\frac{3}{10.12}+\frac{3}{12.14}+.....+\frac{3}{48.50}\)
\(A=\frac{3}{2}.\left(\frac{2}{10.12}+\frac{2}{12.14}+......+\frac{2}{48.50}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{12}+....+\frac{1}{48}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\left(\frac{1}{10}-\frac{1}{50}\right)\)
\(A=\frac{3}{2}.\frac{2}{25}\)
\(A=\frac{3}{25}\)
(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 \(2n⋮n-2\)
\(2n⋮n-2\)
\(\Rightarrow2n-4+4⋮n-2\)
\(\Rightarrow4⋮n-2\)
\(\Rightarrow n-2\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)
| n-2 | -4 | -2 | -1 | 1 | 2 | 4 |
| n | -2 | 0 | 1 | 3 | 4 | 6 |
| Kết luận | loại | t/m | t/m | t/m | t/m | t/m |
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!
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+....+\frac{5}{48.50}\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{5}{2}.\frac{12}{25}=\frac{6}{5}\)
\(\frac{5}{2.4}+\frac{5}{4.6}+\frac{5}{6.8}+...+\frac{5}{48.50}\)
\(=\frac{2}{5}.\left(\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+...+\frac{2}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{4-2}{2.4}+\frac{6-4}{4.6}+\frac{8-6}{6.8}+...+\frac{50-48}{48.50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{48}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\left(\frac{1}{2}-\frac{1}{50}\right)\)
\(=\frac{2}{5}.\frac{12}{25}\)
\(=\frac{24}{125}\)
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
2/5 . A = 2/10.12 + 2/12.14 + ....... + 2/998.1000
= 1/10 - 1/12 + 1/12 - 1/14 + ..... + 1/998 - 1/1000
= 1/10 - 1/1000 = 99/1000
=> A = 99/1000 : 2/5 = 99/400
Tk mk nha
Mình làm tắt cũng dc nhé có gì ko dc hỏi mình nhé
Ta có:A=(5/10-5/12):2+(5/12-2/14):2+...+(5/998-5/1000):2
suy ra A=5/10-5/1000=99/200
Đặt \(A=\frac{2}{10\cdot12}+\frac{2}{12\cdot14}+\frac{2}{14\cdot16}+...+\frac{2}{48\cdot50}\)
\(A=\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
\(A=\frac{1}{10}-\frac{1}{50}=\frac{5}{50}-\frac{1}{50}=\frac{4}{50}=\frac{2}{25}\)
Vậy \(A=\frac{2}{25}\)
= \(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\)
= \(\frac{1}{10}-\frac{1}{50}\)= \(\frac{2}{25}\)