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a) \(\forall\)n \(\in\) N* ta có :
\(\dfrac{1}{n\left(n+1\right)}=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{n+1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) (đpcm)
\(A=\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+...+\dfrac{1}{8}.\dfrac{1}{9}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(=\dfrac{1}{2}-\dfrac{1}{9}\)
\(=\dfrac{7}{18}\)
\(B=\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{110}\)
\(=\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{10.11}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=\dfrac{1}{4}-\dfrac{1}{11}\)
\(=\dfrac{7}{44}\)
Bài 3:
Để A là số nguyên thì \(n-2+5⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;5;-5\right\}\)
hay \(n\in\left\{3;1;7;-3\right\}\)
a) Để A là phân số thì 5 không chia hết cho n-1 hay n-1 không phải Ư(5) mà Ư(5)={-5;-1;1;5}
Ta có bảng sau:
| \(n-1\ne\) | -5 | -1 | 1 | 5 |
| \(n\ne\) | -4 | 0 | 2 | 6 |
Vậy n\(\ne\left\{-4;0;2;6\right\}\)thì A là phân số
n=0 => A=\(\dfrac{5}{0-1}=-5\)
n=10 => A=\(\dfrac{5}{10-1}=\dfrac{5}{9}\)
n=-2 => A=\(\dfrac{5}{-2-1}=-\dfrac{5}{3}\)
Để A là số nguyên =>5 chia hết cho n-1 <=>n-1 là Ư(5)
Từ bảng trên => n={-4;0;2;6} thì A nguyên
b) Do n là Số tự nhiên => n và n+1 là 2 số tự nhiên liên tiếp
=>n và n+1 nguyên tố cùng nhau
=>phân số \(\dfrac{n}{n+1}\)tối giản(dpcm)
c)\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}=1-\dfrac{1}{50}< 1\left(đpcm\right)\)
c) 1/1.2 + 1/2.3 + 1/3.4 + .....+ 1/49.50
= 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + ......+ 1/49 - 1/50
tới bước đây mik làm gọn lại chút nha
= 1/1 - 1/50
=49/50
Suy ra : 49/50 <1 ( điều phải chứng minh )
Câu 1 :
1/n - 1/n + a = a + n/a ( a + n ) = a + n - a/a ( n + a ) = n/a ( a + n )
Câu 2 :
A = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 +.......+ 1/99 - 1/100
= 1/1 - 1/100 = 99/100
a) Để phân số \(\dfrac{3}{n-2}\) là số nguyên thì n - 2 \(⋮\) 3
\(\Rightarrow\) n - 2 \(\in\) Ư(3)
\(\Rightarrow\) n - 2 \(\in\){3; -3; 1;-1}
n \(\in\){5; -1; 3; 2}
c) \(\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{28.29}\)
\(=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{29}-\dfrac{1}{30}\)
\(=\dfrac{1}{3}-\dfrac{1}{30}\)
\(=\dfrac{10}{30}-\dfrac{1}{30}\)
\(=\dfrac{9}{30}\)
=\(\dfrac{3}{10}\)
\(M=\dfrac{8}{3}\cdot\dfrac{2}{5}\cdot\dfrac{3}{8}\cdot10\cdot\dfrac{19}{92}\\ =\dfrac{8\cdot2\cdot3\cdot10\cdot19}{3\cdot5\cdot8\cdot92}\\ =\dfrac{8\cdot2\cdot3\cdot2\cdot5\cdot19}{3\cdot5\cdot8\cdot2\cdot2\cdot23}\\ =\dfrac{19}{23}\)
\(N=\dfrac{5}{7}\cdot\dfrac{5}{11}+\dfrac{5}{7}\cdot\dfrac{2}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\\ =\dfrac{5}{7}\cdot\left(\dfrac{5}{11}+\dfrac{2}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\cdot\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)
\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\\ =\left(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\right)\cdot0\\ =0\)
a, \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{n+a-n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)
Vậy \(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{a}{n\left(n+a\right)}\)
b,
\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100}\)
\(B=\dfrac{5}{1.4}+\dfrac{5}{4.7}+...+\dfrac{5}{100.103}\)
\(3B=\dfrac{5.3}{1.4}+\dfrac{5.3}{4.7}+...+\dfrac{5.3}{100.103}\)
\(3B=5\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\right)\)
\(3B=5\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(3B=5\left(1-\dfrac{1}{103}\right)=5\cdot\dfrac{102}{103}=\dfrac{510}{103}\)
\(B=\dfrac{510}{103}:3=\dfrac{170}{103}\)
\(C=\dfrac{1}{15}+\dfrac{1}{35}+...+\dfrac{1}{2499}\)
\(C=\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{49.51}\)
\(2C=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{49.51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{49}-\dfrac{1}{51}\)
\(2C=\dfrac{1}{3}-\dfrac{1}{51}=\dfrac{16}{51}\)
\(C=\dfrac{16}{51}:2=\dfrac{8}{51}\)
Bài 1:
a) \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Quy đồng \(VP\) ta được:
\(VP=\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow VP=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}\)
\(\Rightarrow VP=\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
\(\Rightarrow VP=VT\)
Vậy \(\forall n\in Z,n>0\Rightarrow\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\) (Đpcm)
b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
Bài 3:
a) \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)
b) A=\(\dfrac{1}{2}.\dfrac{1}{3}+\dfrac{1}{3}.\dfrac{1}{4}+\dfrac{1}{4}.\dfrac{1}{5}+\dfrac{1}{5}.\dfrac{1}{6}+\dfrac{1}{6}.\dfrac{1}{7}+\dfrac{1}{7}.\dfrac{1}{8}+\dfrac{1}{8}.\dfrac{1}{9}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\)
\(=\dfrac{1}{2}-\dfrac{1}{9}\)
\(=\dfrac{7}{18}\)
B=\(\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)
\(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}\)
\(=\dfrac{1}{5}-\dfrac{1}{12}\)
\(=\dfrac{7}{60}\)