Sửa đề: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)

Ta có: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-399}{400}\)

\(=\dfrac{-3\cdot8\cdot15\cdot...\cdot399}{4\cdot9\cdot16\cdot...\cdot400}\)

\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}{2^2\cdot3^2\cdot4^2\cdot...\cdot20^2}\)

\(=\dfrac{-2\cdot3\cdot4\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot20}\)

\(=\dfrac{-1}{20}\cdot\dfrac{21}{2}\)

\(=\dfrac{-21}{40}\)

a) $2020.2020-2022.2018$

$ = 2020^2-(2020+2).(2020-2)$

$ = 2020^2 - (2020^2-2^2)$

$ = 4$

b) \(\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{400}-1\right)\)

\(=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{20^2}-1\right)\)

\(=\dfrac{\left(-1\right)\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot\cdot\cdot\left(-19\right)\cdot21}{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot20^2}\)

\(=-\dfrac{1}{20}\cdot\dfrac{21}{2}=-\dfrac{21}{40}\)

Giải:

a) 

b) =(1/2²−1)(1/3²−1)(1/4²−1)...(1/20²-1)

=−1/20⋅21/2

=-3/4×(-8/9)×(-15/16)×....×(-399/400)

=-3×8×15×....×399/4×9×16×...×400

=-3×2×4×3×5×.....×21×19/2^2×3^2×4^2×....×20^2

=-2×3×4×...×19/2×3×4×...×20 × 3×4×5×....×21/2×3×4×....×20

=-1/20×21/20

=-21/40

\(=\left(1-\dfrac{1}{2}\right)\left(1+\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{20}\right)\left(1+\dfrac{1}{20}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{19}{20}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{21}{20}\)

=1/20*21/2

=21/40