bài 1:Cho b=2+22+23.+.......+2100
Tìm x để 22*x-1=b
bài 2:So sánh:
A=1+2+22+.........+22002 va B=2003
nhờ các bạn giải thật chi tiết giúp tớ nha!!!!!
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\(A=1+2+2^2+...+2^{2022}\)
\(\Rightarrow2A=2+2^2+...+2^{2023}\)
\(\Rightarrow2A-A=2^{2023}-1\)
\(\Rightarrow A=2^{2023}-1\)
\(\Rightarrow A< 2^{2023}=2^2.2^{2021}=4.2^{2021}< 5^{2021}\)
\(\Rightarrow A< B\)
Ta có:
\(A=1+2+2^2+...+2^{2002}\)
\(2A=2+2^2+2^3+...+2^{2003}\)
\(2A-A=\left(2+2^2+2^3+...+2^{2003}\right)-\left(1+2+2^2+....+2^{2002}\right)\)
\(A=2^{2003}-1\)
Mà: \(2^{2003}=2^{2003}\)
\(\Rightarrow2^{2003}-1< 2^{2003}\)
\(\Rightarrow A< B\)
A=1/2+1/22+1/23+...+1/22020+1/22021 > B=1/3+1/4+1/5+13/60
\(a,\Rightarrow2A=2+2^2+...+2^{2011}\)
\(\Rightarrow2A-A=2+2^2+...+2^{2011}-2^0-2-..-2^{2010}\)
\(\Rightarrow A=2^{2011}-1=B\)
\(b,A=2019.2011=\left(2010-1\right)\left(2010+1\right)=\left(2010-1\right).2010+\left(2010-1\right)=2010^2-2010+2010-1=2010^2-1< 2010^2=B\)
\(a,\Rightarrow2A=2^1+2^2+...+2^{2011}\\ \Rightarrow2A-A=A=2^{2011}-2^0=2^{2011}-1=B\)
\(b,A=\left(2010-1\right)\left(2010+1\right)=2010^2+2010-2010-1=2010^2-1< 2010^2=B\)
\(2^{x+3}.2=2^2.3+52\)
\(=>2^{x+3}.2=64\)
\(=>2^{x+3}=64:2\)
\(=>2^{x+3}=32\)
\(=>2^{x+3}=2^5\)
=>x+3=5
=>x=5-3
=>x=2
Vậy ...........
2x + 3 . 2 = 22 . 3 + 52
2x + 3 . 2 = 4 . 3 + 52
2x + 3 . 2 = 12 + 52
2x + 3 . 2 = 64
2x + 3 = 64 : 2
2x + 3 = 32
2x + 3 = 25
x + 3 = 5
x = 5 - 3
x = 2
Vậy x = 2
= 2/5 + 1/5 + 1/5
= 4/5
b. = 2/7 + 4/7 + 5/21
= 6/21 + 12/21 + 5/21
= 23/21
HT
a) \(2^x=16=2^4\Rightarrow x=4\)
b) \(x^3=27=3^3\Rightarrow x=3\)
c) \(x^{50}=x\Rightarrow x\left(x^{49}-1\right)=0\Rightarrow x=0\) hay \(x=1\)
d) \(\left(x-2\right)^2=16=4^2\Rightarrow x-2=4\) hay \(x-2=-4\)
\(\Rightarrow x=6\) hay \(x=-2\)
a) \(2^{300}=2^{3.100}=8^{100}\)
\(3^{200}=3^{2.100}=9^{100}\)
vì \(8^{100}< 9^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
b) \(3^{500}=3^{5.100}=243^{100}\)
\(7^{300}=7^{3.100}=343^{100}\)
vì \(243^{100}< 343^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
\(A=2+2^2+...+2^{2002}\)
\(2A=2^2+2^3+...+2^{2003}\)
\(2A-A=2^2+2^3+...+2^{2003}-2-2^2-...-2^{2002}\)
\(A=2^{2003}-2\)
Mà \(2^{2003}-2< 2^{2003}\Rightarrow A< B\)
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