Question

Tính

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

Answer 1

Cô giải như sau Minh nhé :)

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

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

Answer 2

( 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