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\(A=1-3+3^2-3^3+...+3^{2021}-3^{2022}\)
\(3A=3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\)
\(3A-A=\left(1-3+3^2-3^3+...+3^{2021}-3^{2022}\right)-\left(3-3^2+3^3-3^4+...+3^{2022}-3^{2023}\right)\)
\(2A=3^{2023}-1\)
\(\Rightarrow A=\left(3^{2023}-1\right)\div2\)
\(\text{cái này mình sợ sai nên bạn có thể nhờ cô chữa}\)
Bài 1 :
\(M=\dfrac{30-2^{20}}{2^{18}}=\dfrac{2.15-2^{20}}{2^{18}}=\dfrac{15}{2^{17}}-2^2=\dfrac{15}{2^{17}}-4< 0\left(\dfrac{15}{2^{17}}< 1\right)\)
\(N=\dfrac{3^5}{1^{2021}+2^3}=\dfrac{3^5}{9}=\dfrac{3^5}{3^2}=3^3=27\)
\(\Rightarrow M< N\)
Bài 3 :
a) \(t^2+5t-8\) khi \(t=2\)
\(=5^2+2.5-8\)
\(=25+10-8\)
\(=27\)
b) \(\left(a+b\right)^2-\left(b-a\right)^3+2021\left(1\right)\)
\(\left\{{}\begin{matrix}a=5\\b=a+1=6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=11\\b-a=1\end{matrix}\right.\)
\(\left(1\right)=11^2-1^3+2021=121-1+2021=2141\)
c) \(x^3-3x^2y+3xy^2-y^3=\left(x-y\right)^3\left(1\right)\)
\(\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\) \(\Rightarrow x-y=1\)
\(\left(1\right)=1^3=1\)
`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`
a,( 393+390) : (317. 373)
= (33+1). 390 : 390
= 33+1
=27+1
=28
b,(556+57) : (549+1)
=57. (549+1) : (549+1)
=57= 78125
c,(722+721+720) ; (25+24+32)
= 720. (72+71+1) : [24. (2+1)+32 ]
= 720. 57 : [ 24. 3 +32 ]
= 720. 57 : ( 24+3) . 3
= 720. 57 : 19 . 3
= 720. 57 : 57
= 720
3A-A= 3^2+3^3+....+3^101-3 -3^2-3^3-....-3^100
A= (3^101-3 ) :2
A = 3 + 32 + 33 + .... + 3100
3A = 32 + 33 + 34 + ... + 3101
3A - A = (32 + 33 + 34 + ... + 3101) - (3 + 32 + 33 + ... + 3100)
2A = 3101 - 3
=> A = \(\frac{3^{101}-3}{2}\)
Ủng hộ mk nha !!! ^_^
\(A=1-3+3^2-3^3+...-3^{2021}+3^{2022}\)
\(\Rightarrow3A=3-3^2+3^3-3^4+...-3^{2022}+3^{2023}\)
\(\Rightarrow3A+A=4A\)
\(=\left(1-3+3^2-3^3+...-3^{2021}+3^{2022}\right)+\left(3-3^2+3^3-3^4+...-3^{2022}+3^{2023}\right)\)
\(=1+3^{2023}\)
\(\Rightarrow4A-3^{2023}=1+3^{2023}-3^{2023}=1\)
cảm ơn bạn