Question

[Tô Thị Mai Trang]

^{chứng minh rằng}

^1, 1/n(n+1)=1/n-1/n+1

^2, 2/n(n+1)(n+2)=1/n(n+1)-1/(n+1)(n+2)

^3, 3/n(n+1)(n+2)(n+3)=1/n(n+1)(n+2)-1/(n+1)(n+2)(n+3)

^4,  4/(2n-1)(2n+1)(2n+3)=1/(2n+1)(2n-1)-1/(2n+1)(2n+3)

^5, m/n(n+m)=1/n-1/n+m

^6,  2m/n(n+m)(n+2n)=1/n(n+m)-1/(n+m)(n+2n)

ai nhanh mình tick trước 9 giờ

Answer 1

[나 재민]

\(1) VP= \frac{1}{n}-\frac{1}{n+1}\)\(= \frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}\)\(= \frac{n+1-n}{n(n+1)}\)\(= \frac{1}{n(n+1)}\)\(= VT\)

2) \(VP= \frac{1}{n+1}-\frac{1}{(n+1)(n+2)}= \frac{(n+2)}{n(n+1)(n+2)}-\frac{n}{n(n+1)(n+2)}\)\(= \frac{n+2-n}{n(n+1)(n+2)}= \frac{2}{n(n+1)(n+2)}=VT\)

3) \(VP= \frac{1}{n(n+1)(n+2)}-\frac{1}{(n+1)(n+2)(n+3)}=\frac{n+3}{n(n+1)(n+2)(n+3)}-\frac{n}{n(n+1)(n+2)(n+3)}\)\(= \frac{n+3-n}{n(n+1)(n+2)(n+3)}=\frac{3}{n(n+1)(n+2)(n+3)(n+4)}=VT\)

Những ý sau làm tương tự, thế mà chẳng thèm mở mồm ra hỏi bạn :))

Answer 2

[배 성 달]

chị thương ơi gửi em câu 6,7