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b) 32, 33, 34 , 35.
\(a,2^2=4,2^3=8,2^4=16,2^5=32,2^6=64,2^7=128,2^8=256,2^9=512,2^{10}=1024\)
\(b,3^2=9,3^3=27,3^4=81,3^5=243\)
\(c,4^2=16,4^3=64,4^4=256\)
\(d,5^2=25,5^3=125,5^4=625\)
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
a) \(-\frac{11}{18}\)
b)\(-\frac{3}{2}\)
c)\(\frac{49}{78}\)
d)\(\frac{23}{11}\)
e) \(\frac{11.12+22.24+44.48}{33.36+66.72+132.144}\)
\(=\frac{11.12+22.24+44.48}{11.3.12.3+22.3.3.24+44.3.348}\)
\(=\frac{11.12+22.24+44.48}{\left(1.12+22.24+44.48\right).9}\)
\(=\frac{1}{9}\)
a) A = 2 + 22 + 23 +...+ 229
=>2A= 22 + 23 +...+ 230
=>2A-A= 22 + 23 +...+ 230-2-22-23-...-229
=>A.(2-1)=230-2
=>A=230-2
b) B = 1 + 3 + 32 + 33 + ... + 339
=>3B=3 + 32 + 33 + ... + 340
=>3B-B=3 + 32 + 33 + ... + 340-1 - 3 - 32 - 33 - ... - 339
=>B(3-1)=340-1
=>B.2=340-1
=>B=\(\frac{3^{40}-1}{2}\)
nhiều wa 2 câu trước
a) A= 2 + 2^2 + 2^3 + ... + 2^29
2A= 2. (2 + 2^2 + 2^3 + ... + 2^29)
2A= 2^2 + 2^3 + 2^4 + ... +2^29 + 2^30
- (dấu trừ viết ra đầu dòng nha)
A= 2 + 2^2 + 2^3 + 2^4 ... + 2^29
1A= 2^29 - 2
A= 2^29 -2 trên 1
kick mk nha