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a) 18 2 < 10 3
b) 3 2 + 4 2 < ( 3 + 4 ) 2
c) 100 2 + 30 2 < ( 100 + 30 ) 2
d) a 2 + b 2 > ( a - b ) 2 với a ∈ N * ; b ∈ N * .
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Bài 1:
a. $2^{29}< 5^{29}< 5^{39}$
$\Rightarrow A< B$
b.
$B=(3^1+3^2)+(3^3+3^4)+(3^5+3^6)+...+(3^{2009}+3^{2010})$
$=3(1+3)+3^3(1+3)+3^5(1+3)+...+3^{2009}(1+3)$
$=(1+3)(3+3^3+3^5+...+3^{2009})$
$=4(3+3^3+3^5+...+3^{2009})\vdots 4$
Mặt khác:
$B=(3+3^2+3^3)+(3^4+3^5+3^6)+....+(3^{2008}+3^{2009}+3^{2010})$
$=3(1+3+3^2)+3^4(1+3+3^2)+...+3^{2008}(1+3+3^2)$
$=(1+3+3^2)(3+3^4+....+3^{2008})=13(3+3^4+...+3^{2008})\vdots 13$
Bài 1:
c.
$A=1-3+3^2-3^3+3^4-...+3^{98}-3^{99}+3^{100}$
$3A=3-3^2+3^3-3^4+3^5-...+3^{99}-3^{100}+3^{101}$
$\Rightarrow A+3A=3^{101}+1$
$\Rightarrow 4A=3^{101}+1$
$\Rightarrow A=\frac{3^{101}+1}{4}$
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2^5=32
5^3=125 , 3^5=143
=)3^5>5^3
14^12- 14^9=2744
14^3=2744
(n^54)^2=n^108
n5=32
n-2
so sánh :53 và 35
Ta có 53=125
35=243
=>125<243
hay 53<35
(n54) 2 = n108
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Ta có: 32n = (32)n = 9n
23n = (23)n = 8n
Vì 9 > 8 => 9n > 8n => 32n > 23n
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Ta có: \(\dfrac{n+1}{n+5}-\dfrac{n+2}{n+3}\)
\(=\dfrac{n^2+4n+3-n^2-7n-10}{\left(n+5\right)\left(n+3\right)}\)
\(=\dfrac{-3n-7}{\left(n+5\right)\left(n+3\right)}\)
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\(M=\frac{10^{30}+1}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10\cdot(10^{30}+1)}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+10}{10^{31}+1}\)
\(\Rightarrow10M=\frac{10^{31}+1+9}{10^{31}+1}\)
\(\Rightarrow10M=1+\frac{9}{10^{31}+1}\)
\(N=\frac{10^{31}+1}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10\cdot(10^{31}+1)}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+10}{10^{32}+1}\)
\(\Rightarrow10N=\frac{10^{32}+1+9}{10^{32}+1}\)
\(\Rightarrow10N=1+\frac{9}{10^{32}+1}\)
Mà\(1+\frac{9}{10^{31}+1}>1+\frac{9}{10^{32}+1}\)
Nên \(10M>10N\)
Hay \(M>N\)
3^2.n và 2^3.n
3^2.n=(3^2)^n=9^n
2^3.n=(2^3)^n=8^n
Vì 9^n>8^n=>3^2.n>2^3.n
Vậy 3^2.n>2^3.n