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\(8^n:2^n=16^{2011}\)
\(\left(2^3\right)^n:2^n=\left(2^4\right)^{2011}\)
\(2^{3n}:2^n=2^{8044}\)
\(2^{3n-n}=2^{8044}\)
\(\Rightarrow3n-n=8044\)
\(2n=8044\)
\(\Rightarrow n=\frac{8044}{2}\)
\(n=4022\)
Vậy \(n=4022\)
5300 = (53)100 = 125100
3500= (35)100= 243100
Vì 125100 < 243100 nên 5300 < 3500
Vậy...
2 mũ 30=6 mũ 10
3 mũ 30=9 mũ 10
4 mũ 30=12 mũ 10
6 mũ 10 + 9 mũ 10 + 12 mũ 10 =6.9.12 mũ 10 > 3 . 24 mũ 10
**** e nha chị moon
\(3^{300}+4^{300}\)
\(=27^{100}.64^{100}\)
\(=1728^{100}>3.24^{100}\)
\(S=1+5+5^2+5^4+...+5^{200}\)
\(\Leftrightarrow5^2S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S=5^2+5^4+...+5^{202}\)
\(\Leftrightarrow25S-S=5^{202}-1\)
\(\Leftrightarrow S=\left(5^{202}-1\right)\div24\)
a) S = 1 + 52 + 54 + ... + 5200
=> 52S = 52.(1 + 52 + 54 + ... + 5200)
=> 25S = 52 + 54 + 56 + ... + 5202
=> 25S - S = (52 + 54 + 56 + ... + 5202) - (1 + 52 + 54 + ... + 5200)
=> 24S = 5202 - 1
=> S = \(\frac{5^{202}-1}{24}\)
Bạn tham khảo nhé
a ) Ta có :
\(\left(-\frac{1}{5}\right)^{300}=\left(\frac{1}{5}\right)^{300}=\frac{1}{5^{300}}=\frac{1}{\left(5^3\right)^{100}}=\frac{1}{125^{100}}\)
\(\left(-\frac{1}{3}\right)^{500}=\left(\frac{1}{3}\right)^{500}=\frac{1}{3^{500}}=\frac{1}{\left(3^5\right)^{100}}=\frac{1}{243^{100}}\)
Do \(\frac{1}{125^{100}}>\frac{1}{243^{100}}\left(125^{100}< 243^{100}\right)\)
\(\Rightarrow\left(-\frac{1}{5}\right)^{300}>\left(-\frac{1}{3}\right)^{500}\)
b )
Ta có :
\(2550^{10}=\left(50.51\right)^{10}=50^{10}.51^{10}\)
\(50^{20}=50^{10}.50^{10}\)
Do \(50^{10}.51^{10}>50^{10}.50^{10}\)
\(\Rightarrow50^{20}< 2550^{10}\)
c )
Ta có :
\(2^{100}=\left(2^4\right)^{25}=16^{25}\)
\(3^{75}=\left(3^3\right)^{25}=27^{25}\)
\(5^{50}=\left(5^2\right)^{25}=25^{25}\)
Do \(16^{25}< 25^{25}< 27^{25}\)
\(\Rightarrow2^{100}< 5^{50}< 3^{75}\)
+)\(8^2=\left(2^3\right)^2=2^6\)
+)\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9>8\Rightarrow9^{100}>8^{100}\)hay \(3^{200}>2^{300}\)
+)\(9^{20}=\left(3^2\right)^{20}=3^{40}\)
\(27^{13}=\left(3^3\right)^{13}=3^{39}\)
Vì \(40>39\Rightarrow3^{40}>3^{39}\)hay \(9^{20}>27^{13}\)
+)\(10^{20}=10^{2.10}=\left(10^2\right)^{10}=100^{10}\)
\(2^{100}=2^{10.10}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì \(100< 1024\Rightarrow100^{10}< 1024^{10}\)hay \(10^{20}< 2^{100}\)
+)\(2^{161}=2^{4.40+1}=\left(2^4\right)^{40}.2=16^{40}.2\)
Vì \(13< 16\Rightarrow13^{40}< 16^{40}\)\(\Rightarrow13^{40}< 2^{161}\)
Ta có:
2300=(23)100=8100
3200=(32)100=9100
Vì 8<9 nên 8100<9100
Vậy 2300<3200
Suy ra A<B
Ta có:
\(A=2^{300}\)\(=\left(2^3\right)^{100}\)\(=8^{100}\)
\(B=3^{200}\)\(=\)\(\left(3^2\right)^{100}\)\(=9^{100}\)
Vì \(8^{100}< 9^{100}\)nên \(A< B\)
\(a)\)
Cách 1 :
\(2^{30}+3^{30}+4^{30}\ge3\sqrt[3]{\left(2.3.4\right)^{30}}=3.\left(2.3.4\right)^{10}=3.24^{10}\) ( Cosi )
Mà \(2^{30}\ne3^{30}\ne4^{30}\) nên dấu "=" không xảy ra hay \(2^{30}+3^{30}+4^{30}>3.24^{10}\)
Vậy ...
Cách 2 :
\(4^{30}=4^{11}.4^{19}=4^{11}.2^{38}>3^{11}.2^{30}=3.3^{10}.8^{10}=3.24^{10}\)
Vậy ...
\(b)\)\(4+\sqrt{33}=\sqrt{16}+\sqrt{33}>\sqrt{14}+\sqrt{29}\)
Vậy ...
Lớn hơn nhé !
Ta có :
\(3.24^{100}=3.3^{100}.8^{100}=3^{101}.8^{100}\)
Xét : \(4^{300}\)và \(3^{101}.8^{100}\)ta có :
\(4^{300}=2^{300}.2^{300}=\left(2^2\right)^{150}.\left(2^3\right)^{100}=\)\(4^{150}.8^{100}\)
Vì \(8^{100}=8^{100}\)và \(4^{150}>3^{101}\Rightarrow4^{300}>3^{101}.8^{100}\)
\(\Rightarrow4^{300}+3^{400}>3.24^{100}\)