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Bài 2:
Ta có: \(16x+40=10\cdot3^2+5\left(1+2+3\right)\)
\(\Leftrightarrow16x+40=90+30\)
\(\Leftrightarrow16x=80\)
hay x=5
Bài 1: Bạn xem lại đã viết đúng đề chưa vậy.
Bài 2:
$P=29-|16+3.2|+1=29-|22|+1=29-22+1=7+1=8$
\(\left(\frac{1}{2}\right)^5.2^5-\left(\frac{1}{2019}-\frac{1}{2020}+\frac{1}{2021}\right)\)
\(=\frac{1^5}{2^5}.2^5-\left(\frac{1}{2019}-\frac{1}{2020}+\frac{1}{2021}\right)\)
\(=\frac{1^5.2^5}{2^5}-\left(\frac{2020.2021}{2019.2020.2021}-\frac{2019.2021}{2019.2020.2021}+\frac{2019.2020}{2019.2020.2021}\right)\)
\(=1^5-\left(\frac{2020.2021-2019.2021+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{\left(2020-2019\right).2021+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1.2021+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1+2020+2019.2020}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1+2020.\left(1+2019\right)}{2019.2020.2021}\right)\)
\(=1-\left(\frac{1+2020.2020}{2019.2020.2021}\right)\)
\(=1-\frac{1+2020}{2019.2021}\)
\(=1-\frac{2021}{2019.2021}\)
\(=1-\frac{1}{2019}\)
\(=\frac{2019}{2019}-\frac{1}{2019}=\frac{2018}{2019}\)
Chúc bạn học tốt
a) S=1 + 2 + 2^2 + 2^3 +...+ 2^63
2S=2 + 2^2 + 2^3 + 2^4 +...+ 2^64
S=2S-S=(2 + 2^2 + 2^3 + 6^4 +...+ 2^64)-(1 + 2 + 2^2 + 2^3 +...+ 2^63)
S=2 + 2^2 + 2^3 + 2^4 +...+ 2^64 - 1 - 2 - 2^2 - 2^3 -...- 2^63
S=2^64 - 1
\(S=1+2+2^2+...+2^{2021}\)
=>\(2S=2+2^2+2^3+...+2^{2022}\)
=>\(2S-S=2+2^2+...+2^{2022}-1-2-2^2-...-2^{2021}\)
=>\(S=2^{2022}-1\)
S=1+2+2^2+2^3+...+2^2022
2S=2+2^2+2^3+2^4+...+2^2023
2S-S=S=2^2023-1/2