Question

a=2020. 2020-2022.2018

b= (1/4 -1).(1/9-1).(1/16-1)....(1/400-1)

Answer 1

a) $2020.2020-2022.2018$

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

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

$ = 4$

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

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

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

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

Answer 2

Giải:

a) 

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

=−1/20⋅21/2