Đặt \(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)\left(1+\frac{7}{48}\right)+...+\left(1+\frac{7}{2009}\right)\)

\(\Leftrightarrow1+\left(\frac{7}{9}.\frac{7}{20}.\frac{7}{33}.\frac{7}{48}.....\frac{7}{2009}\right)\)

Dãy phân số trên có số phân số là:

(2009 - 9) : 4 + 2 =502

\(\Rightarrow A=1+\left(\frac{7^{502}}{9.20.33.48.....2009}\right)\)

giúp mình với

A=16/9 .27/20 . 40/33 . 55/48 ....2016/2009

   =(16.27.40.55...2016)  /  (9.20.33.48...2009)

   = [(2.8)(3.9)(4.10)(5.11)...(42.48)]   /  [(1.9)(2.10)(3.11)(4.12)...(41.49)]

   =[(2.3.4.5..42)(8.9.10.11..48)]   /   [(1.2.3.4...41)(9.10.11.12...49)]

   =(42.8) / (1.49)

   =336/49

  =48/7

Cậu thông cảm,mk trình bày hơi khó nhìn

Cô giải như sau Minh nhé :)

\(A=\left(1+\frac{7}{9}\right)\left(1+\frac{7}{20}\right)\left(1+\frac{7}{33}\right)...\left(1+\frac{7}{2900}\right)=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}...\frac{2907}{2900}\)

\(=\frac{8.2}{9.1}.\frac{9.3}{10.2}.\frac{10.4}{11.3}....\frac{57.51}{58.50}=\frac{\left(8.9.10....57\right)\left(2.3.4...51\right)}{\left(9.10.11...58\right)\left(2.3.4....50\right)}=\frac{8.51}{58}=\frac{204}{29}\)

( 1 + 7/9 ) x ( 1 + 7/20 ) x ( 1 + 7/33 ) x...x ( 1 + 7/2900)

= (8x2)/(9x1) x (9x3)/(10x2) x (10x4)/(11x3) x...x (57x51)(58x50)

=(8x2x9x3x10x4x...x57x51) / (9x1x10x2x11x3x...x58x50) Sau khi giản ước ta được :

= (8x51) / (1x58) = 204/29

\(A=\frac{16}{9}.\frac{27}{20}.\frac{40}{33}.......\frac{2907}{2900}\)

\(A=\frac{2.8}{1.9}.\frac{3.9}{2.10}.\frac{4.10}{3.11}......\frac{51.57}{50.58}\)

\(A=\frac{2.3.4.....51}{1.2.3...50}.\frac{8.9.10....57}{9.10.11...58}\)

\(A=51.\frac{8}{58}=\frac{204}{29}\)

Bạn Nguyễn Tuấn Minh làm đúng rùi đó !!! Chuẩn ý kiến mk...^.^