[120:3^(3)*2-50:2-(3^(2)-2.4)^(2021)]}
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a) 17.13+17.42-17.35
=17.(13+42-35)
=17.20=340
b) [25.(18-42)-10]:4+6
=(25.2-10):4+6
=40:4+6=16
c) 36:32+23.22-32.3
=34+25-33
=81+32-27=86
d) B=3.42-22.3
=3.(16-4)
=3.12=36
e)20220+3.[52.10-(23-13)2]
=1+3.(250-100)
=1+450=451
g) 27.77+24.27-27
=27.(77+24-1)
=27.100=2700
h) 5.23+79:77-12020
=40+72-1
=89-1=88
i) 120:{54[50:2+(32-2.4)]}
=120:[54(25+1)]
=120:1404=10/117
(chỉnh đề)
A=\(-1+2-3-4-5+6-7-8-9+...-2021-2022+2023-2024\)
=\(\left(-1-2024\right)+\left(2+2023\right)+\left(-3-2022\right)+\left(-4-2021\right)+\left(-5-2020\right)+\left(6+2019\right)-\left(-7-2018\right)+\left(-8-2017\right)+\left(-9-2016\right)+...+\left(1010+1015\right)+\left(-1011-1014\right)+\left(-1012-1013\right)\)=\(-2025+2025-2025-2025-2025+2025-2025-2025-2025+...+2025-2025-2025\)=253.2025-1771.2025=-3 073 950.
B=\(1.3.5+3.5.7+5.7.9+7.9.11+...+99.101.103\)
8B=\(1.3.5.8+3.5.7.8+5.7.9.8+7.9.11.8+...+99.101.103.8\)
8B=\(1.3.5.\left[7-\left(-1\right)\right]+3.5.7.\left(9-1\right)+5.7.9.\left(11-3\right)+7.9.11.\left(13-5\right)+...+99.101.103.\left(105-97\right)\)8B=\(3.5+3.5.7+3.5.7.9-3.5.7+5.7.9.11-3.5.7.9+7.9.11.13-5.7.9.11+...+99.101.103.105-97.99.101.103\)
B=\(\dfrac{3.5+99.101.103.105}{8}=13517400\)
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
\(\text{= 2/1 . 2/3 . 3/2 . 3/4 . 4/3 . 4/5 ....... 50/49.50/51 }\)
Dùng phương pháp khử liên tiếp ta có
\(=\frac{2}{1}-\frac{50}{51}=\frac{52}{51}\)