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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
\(3^{600}=3^{200.3}=\left(3^3\right)^{200}=9^{200}^{_{\left(1\right)}}\)
\(4^{400}=\left(2^2\right)^{400}=2^{800}=2^{200.4}=\left(2^4\right)^{200}=16^{200}_{\left(2\right)}.\)
\(\left(1\right),\left(2\right)\Rightarrow4^{400}>3^{600}\)
\(4^{32}=\left(2^2\right)^{32}=2^{64}_{\left(1\right)}\)
\(16^{15}=\left(2^4\right)^{15}=2^{60}_{\left(2\right)}\)
\(\left(1\right),\left(2\right)\Rightarrow4^{32}>16^{15}\)
\(2^{91}=\left(2^{13}\right)^7=73728^7\)
\(5^{35}=\left(5^5\right)^7=3125^7\) nhỏ hơn \(73728^7\)
\(\Rightarrow2^{91}\) lớn hơn \(5^{35}\)
\(b,3^{400}=\left(3^4\right)^{100}=81^{100}\\ 4^{300}=\left(4^3\right)^{100}=64^{100}\\ Vì:81^{100}>64^{100}\left(Do:81>64\right)\\ \Rightarrow3^{400}>4^{300}\)
5^200 = (5^2)^100=25^100
2^400 = (2^4)^100=8^100
Mà 25^100>8^100
=> 5^200 > 2^400
ta có : \(3^{600}=\left(3^3\right)^{200}=9^{200}\)
\(5^{400}=\left(5^2\right)^{200}=25^{200}\)
vì \(9< 25\Rightarrow3^{600}< 5^{400}\)
Ta có:
+ 3600= 33.200 =(33)200=9200
+5400=52.200 =(52)200=25200
Do 9< 25 nên 9200 < 25200
Vậy 3600< 5400
\(3^{600};4^{400}\)
\(3^{600}=\left(3^3\right)^{200}\)
\(4^{400}=\left(4^2\right)^{200}\)
Vì : \(27^{200}>16^{200}\)
\(\Rightarrow3^{600}>4^{400}\)
Ta có:
\(3^{600}=3^{3\times200}=\left(3^3\right)^{200}=27^{200}\)
\(4^{400}=4^{2\times200}=\left(4^2\right)^{200}=16^{200}\)
Vì 27 > 16 \(\Rightarrow27^{200}>16^{200}\Leftrightarrow3^{600}>4^{400}\)
Bài này dễ mà bạn cũng hỏi =(((
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)....\left(\frac{1}{400}-1\right)\)
\(\Leftrightarrow A=\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}....\frac{-399}{400}\)
\(=\frac{1.\left(-3\right)}{2.2}.\frac{2.\left(-4\right)}{3.3}.\frac{3.\left(-5\right)}{4.4}....\frac{19.\left(-21\right)}{20.20}\)
\(=\frac{\left(1.2.3...19\right).\left(\left(-3\right).\left(-4\right).\left(-5\right)...\left(-21\right)\right)}{\left(2.3.4...20\right)\left(2.3.4...20\right)}=\frac{1}{20}.\frac{\left(-21\right)}{2}=\frac{-21}{40}\)
Dễ dàng nhận thấy \(\frac{21}{40}>\frac{1}{2}\Rightarrow\frac{-21}{40}< \frac{-1}{2}\)
Vậy \(A< -\frac{1}{2}\)
\(4^{400}+4^{399}>5^{400}\)