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b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)
\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)
\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)
\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)
\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)
\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)
=2745427451
c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)
\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)
=0+9
=9
b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)
\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)
\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)
\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)
\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)
\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)
=2745427451
c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)
\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)
=0+9
=9
c) 3⁴ - 5.42
= 81 - 210
= -129
d) 120 : {54 - [50 : 2 - (3² - 2.4)]}
= 120 : {54 - [25 - (9 - 8)]}
= 120 : [54 - (25 - 1)]
= 120 : (54 - 24)
= 120 : 30
= 4
8 − 9 4 + 2 7 − − 6 − 3 7 + 5 4 − 3 + 2 4 − 9 7 = 8 − 9 4 + 2 7 + 6 + 3 7 − 5 4 − 3 − 2 4 + 9 7 = 8 + 6 − 3 − 9 4 − 5 4 − 2 4 + 2 7 + 3 7 + 9 7 = 11 − 4 + 2 = 9
81 : 3^2 + 5 x 3
= 81 : 9 + 5 x 3
= 9 + 5 x 3
= 9 + 15
= 24
81 : 3^2 + 5 x 3 = 3^4 : 3^2 + 5 x 3
= 3^2 + 15 = 9 + 15 = 24
= 280 : [25 + ( 30 - 3 mũ 3 ) ]
= 280 : [ 25 + ( 30 - 27 ) ]
= 280 : ( 25 + 3 )
= 280 : 28
=10