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a/ 35 – 2.1111 + 3. 72
= 35 - 2.1 + 3.49
= 35 - 2 + 147
= 33 + 147
= 180
b/ 5.43 + 2.3 – 81. 2 + 7
= 5.64 + 6 - 162 + 7
= 320 + 6 - 162 + 7
= 326 - 162 + 7
= 164 + 7
= 171
c/ 25 + 2.{12 + 2.[3.(5-2)+1]+1}+1
= 32 + 2.{12 + 2.[3.3+1]+1}+1
= 32 + 2.{12 + 2.10+1}+1
= 32 + 2.{12 + 20 + 1}+1
= 32 + 2.{32+1}+1
= 32 + 2.33 + 1
= 32 + 66 + 1
= 98 + 1
= 99
d/ 17.131 + 69.17
= 17 . (131+69)
= 17 . 200
= 3400
e/ 50 – [30 - (6 - 2)2 ]
= 50 - [30 - 42]
= 50 - 14
= 36
f/ 199 +36 + 201 + 184 + 37
= (199+201)+(36+184)+37
= 400 + 220 + 37
= 657
a) 32 . 53 + 92 = 9 . 125 + 81
= 1 125 + 81 = 1 206
b) 83 : 42 - 52 = 512 : 16 - 25 = 32 - 25 = 7
c) 33 . 92 - 52.9 + 18 : 6 = 27 . 81 - 25 . 9 + 3
= 2 187 - 225 + 3 = 1 962 + 3 = 1 965
=(1-2-3+4)+(5-6-7+8)+...+(2017-2018-2019+2020)+2021-2022-2023
=0+0+...+0-1-2023
=-2024
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}\)
\(\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}\)
\(\Rightarrow A=2-\dfrac{1}{2^{2012}}\)
\(A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A=2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\)
⇒\(2A-A=(2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\))\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)\)
⇒\(A=2-\)\(\dfrac{1}{2^{2012}}\)
a) (-12). (7 - 72) - 25. (55 - 43)
= (-12). (- 65) - 25. 12
= 12. 65 – 12. 25
= 12. (65 - 25)
= 12. 40
= 480
b) (39 - 19) : (- 2) + (34 - 22). 5
= 20 : (- 2) + 12. 5
= - 10 + 60
= 60 - 10
= 50.
a) 32 - 6 . (8 - 23) + 18 = 32 - 6 . (8 - 8) + 18
= 32 - 6 . 0 + 18 = 32 + 18 = 50
b) (3 . 5 - 9)3 . (1 + 2 . 3)2 + 42
= (15 - 9)3 . (1 + 6)2 + 42
= 63 . 72 + 42 = 216 . 49 + 16 = 10 584 + 16 = 10 600
`A = 2 + 2^2+ ... + 2^2017`
`=> 2A = 2^2 + 2^3 + ... + 2^2018`
`=> 2A - A = (2^2 + 2^3 + ... + 2^2018) - (2 + 2^2 + ... +2^2017)`
`=> A = 2^2018 - 2`
`B = 1 + 3^2 + ... + 3^2018`
`=> 3^2B = 3^2 + 3^4 + ... + 3^2020`
`=> 9B-B =(3^2 + 3^4 + ... + 3^2020) - (1 + 3^2 + ... + 3^2018`
`=> 8B = 3^2020 - 1`
`=> B = (3^2020 - 1)/8`
`C = 5 + 5^2 - 5^3 + ... + 5^2018`
`=> 5C = 5^2 + 5^3 - 5^4 + ... +5^2019`
`=> 5C + C = ( 5^2 + 5^3 - 5^4 + ... 5^2019) + (5 + 5^2 - 5^3 + ... + 5^2018)`
`=> 6C = 55 + 5^2019`
`=> C = (5^2019 + 55)/6`
đáp án
a) 1000000
b) 98
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