Question

Tính:

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{10^2}\right)\)

Các CTV giải giúp mình với nha.

Answer 1

Ta có : 

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{10^2}\right)\)

\(=\)\(\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}.....\frac{10^2-1}{10^2}\)

\(=\)\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{9.11}{10.10}\)

\(=\)\(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4.....10.10}\)

\(=\)\(\frac{\left(1.2.3.....9\right).\left(3.4.5.....11\right)}{\left(2.3.4.....10\right).\left(2.3.4.....10\right)}\)

\(=\)\(\frac{11}{2.10}\)

\(=\)\(\frac{11}{20}\)

Chúc bạn học tốt ~ 

Answer 2

(1-1/2^2)(1-1/3^2)(1-1/4^2)....(1-1/10^2)

=3/4.8/9.15/16...99/100

Từ đó tính kết quả là ok

Answer 3

\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)..\left(1-\frac{1}{10^2}\right)\)

\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{99}{100}\)=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{9.11}{10.10}\)

\(=\frac{1.2.3...9}{2.3.4...10}.\frac{3.4.5...11}{2.3.4...10}\)\(=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}\)