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Bài 3
\(\frac{n+6}{n+1}=\frac{n+1+5}{n+1}=\frac{n+1}{n+1}+\frac{5}{n+1}\)
\(=1+\frac{5}{n+1}\)
Vậy để \(\frac{n+6}{n+1}\in Z\Rightarrow1+\frac{5}{n+1}\in Z\)
Hay \(\frac{5}{n+1}\in Z\)\(\Rightarrow n+1\inƯ_5\)
\(Ư_5=\left\{1;-1;5;-5\right\}\)
* \(n+1=1\Rightarrow n=0\)
* \(n+1=-1\Rightarrow n=-2\)
* \(n+1=5\Rightarrow n=4\)
* \(n+1=-5\Rightarrow n=-6\)
Vậy \(n\in\left\{0;-2;4;-6\right\}\)
Bài 2:
\(\frac{10}{3.8}+\frac{10}{8.13}+\frac{10}{13.18}+\frac{10}{18.23}+\frac{10}{23.28}=2\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{23}-\frac{1}{28}\right)\\ =2\left(\frac{1}{3}-\frac{1}{28}\right)\\ =2.\frac{56}{84}\\ =\frac{56}{42}=\frac{28}{21}\)
\(A=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{1}{2018}\)
\(A=1+\left(1+\frac{2017}{2}\right)+\left(1+\frac{2016}{3}\right)+...+\left(1+\frac{1}{2018}\right)\)
\(A=\frac{2019}{2019}+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2018}\)
\(A=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)\)
Ta có: \(\frac{A}{B}=\frac{2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}+\frac{1}{2019}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2019}}=2019\)
Cau a : \(A=\frac{a^2\left(a+2\right)-1}{a^2\left(a+2\right)+2a-1}\) => \(A=\frac{-1}{2a-1}\)
Cau b: Neu a la so nguyen thi 2a -1 chac chan phai chia het cho 1 , con tu so la -1 thi da chia het cho 1 roi => day la phan so toi gian
a) Các số có dạng : \(\frac{1}{a\left(a+1\right)}=\frac{\left(a+1\right)-a}{a\left(a+1\right)}=\frac{1}{a}-\)\(\frac{1}{a+1}\)
Thế vào bởi các số sẽ có kết quả
b) Các số có dạng : \(\frac{1}{a\left(a+2\right)}=\frac{1}{2}.\frac{2}{a\left(a+2\right)}=\frac{1}{2}.\frac{\left(a+2\right)-a}{a\left(a+2\right)}\)\(=\frac{1}{2}.\left(\frac{1}{a}-\frac{1}{a+2}\right)\)
Làm tương tự trên
c) Lấy nhân tử chung là 5 rồi làm như câu a)
Câu 1
a) A=2018!.(2019 - 1 -2018)
=2018!.0
= 0
vậy A= 0
b)\(B=\left(1-\frac{1}{9}+1-\frac{2}{10}+1+\frac{3}{11}+...+1-\frac{150}{158}\right):\left(\frac{1}{4}.\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+...+\frac{1}{158}\right)\right)\)
\(=\left(\frac{8}{9}+\frac{8}{10}+...+\frac{8}{158}\right):\left(\frac{1}{4}\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right)\right)\)
\(=8.\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right):\left(\frac{1}{4}\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{158}\right)\right)\)
\(=8:\frac{1}{4}\)
=32
Vậy B= 32