So sánh
a)\(3^{600}và4^{400}\)
b)\(4^{32}và16^{15}\)
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Ta có:a)\(^{3^{600}}\)=\(^{\left(3^3\right)^{200}}\)=\(^{27^{200}}\) \(^{4^{400}}\)=\(^{\left(4^2\right)^{200}}\)=\(^{16^{200}}\)
vì 27^200>16^200 => 3^600>4^400
b) \(^{4^{32}=4^{2.16}=16^{16}}\) vì 16^16>16^15 => 4^32>16^15
\(3^{600}=3^{200.3}=\left(3^3\right)^{200}=9^{200}^{_{\left(1\right)}}\)
\(4^{400}=\left(2^2\right)^{400}=2^{800}=2^{200.4}=\left(2^4\right)^{200}=16^{200}_{\left(2\right)}.\)
\(\left(1\right),\left(2\right)\Rightarrow4^{400}>3^{600}\)
\(4^{32}=\left(2^2\right)^{32}=2^{64}_{\left(1\right)}\)
\(16^{15}=\left(2^4\right)^{15}=2^{60}_{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow4^{32}>16^{15}\)
a)Ta có : \(32^{10}=2^{50}\)
\(16^{15}=2^{60}\)
Vì 50<60 =>2^50<2^60=>32^10<16^15
Vậy 32^10>16^15
b)Ta có : 6*5^22=(5+1)*5^22=5^23+5^22
Vì 5^23+5^22>5^23=>6*5^22>5^23
Vậy6*5^22>5^23
a/ \(3^{600}=\left(3^3\right)^{200}=\left(27\right)^{200}\)
\(4^{400}=\left(4^2\right)^{200}=\left(16\right)^{200}\)
\(\Leftrightarrow3^{600}>4^{400}\)
b/ \(4^{32}\)
\(16^{15}=\left(4^2\right)^{15}=4^{30}\)
\(\Leftrightarrow4^{32}>16^{15}\)
a)\(3^{600}\) = \(\left(3^3\right)^{200}\) = \(27^{200}\)
\(4^{400}\) = \(\left(4^2\right)^{200}\) = \(16^{200}\)
Vì \(27>16\Rightarrow27^{200}>16^{200}=3^{600}>4^{400}\)
Vậy\(3^{600}>4^{400}\)
b) \(32^{10}=\left(2^5\right)^{10}=2^{50}
\)
\(16^{15}=\left(2^4\right)^{15}=2^{60}\)
Vì \(50< 60\Rightarrow2^{50}< 2^{60}\Rightarrow32^{10}< 16^{15}\)
Vậy\(32^{10}< 16^{15}\)
) 2 x 6 = 12 3 x 7 = 21
12 : 2 = 6 21 : 3 = 7
12 : 6 = 2 21 : 7 = 3
4 x 8 = 32 5 x 9 = 45
32 : 4 = 8 45 : 5 = 9
32 : 8 = 4 45 : 9 = 5
b) 600 : 3 = 200 800 : 4 = 200 400 : 2 = 200
600 : 2 = 300 800 : 2 = 400 500 : 5 = 100
a) 2x6=12 3x7=21
12:2=6 21:3=7
12:6=2 21:7=3
4x8=32 5x9=45
32:4=8 45:5=9
32:8=4 45:9=5
b) 600:3=200 800:4=200 400:2=200
600:2=300 800:2=400 500:5=100
A Legend Never Dies
\(32^{15}=\left(2^5\right)^{15}=2^{5.15}=2^{75}\)
\(4^{39}=\left(2^2\right)^{39}=2^{2.39}=2^{78}\)
Do \(2^{78}>2^{75}\)
\(\Rightarrow4^{39}>32^{15}\)
\(\Rightarrow1+4+4^2+...+4^{39}>32^{15}\)
\(\Rightarrow3\left(1+4+4^2+...+4^{39}\right)>32^{15}\)
Vậy \(A>B\)