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) 2020.2020−2022.2018
=20202−(2020+2).(2020−2)
=20202−(20202−22)
=4
b) =(1/22−1)(1/32−1)(1/42−1)...(1/202-1)
=−1/20⋅21/2
