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Lời giải:
a) $A-B=99.10^k-10^{k+2}-10^k=99.10^k-100.10^k-10^k$
$=10^k(99-100-1)=-2.10^k< 0$
$\Rightarrow A<b$
b) $99^{20}-9999^{10}=99^{20}-(99.101)^{10}$
$<99^{20}-(99.99)^{10}=99^{20}-99^{20}=0$
$\Rightarrow 99^{20}<9999^{10}$
a) \(2^{300}=\left(2^3\right)^{100}=8^{100}\)
\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)
\(\Rightarrow2^{300}< 3^{200}\)
b) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
c) \(3^{500}=\left(3^5\right)^{100}=243^{100}\)
\(7^{300}=\left(7^3\right)^{100}=343^{100}>243^{100}\)
\(\Rightarrow3^{500}< 7^{300}\)
\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)
Bài 1:
a) \(\Rightarrow2^x=2^4\Rightarrow x=4\)
b) \(\Rightarrow3^{x+1}=3^{2x}\Rightarrow x+1=2x\Rightarrow x=1\)
c) \(\Rightarrow2^{3x+2}=2^{2x+10}\Rightarrow3x+2=2x+10\Rightarrow x=8\)
d) \(\Rightarrow3^{2x-1}=3^5\Rightarrow2x-1=5\Rightarrow x=3\)
Bài 2:
a) \(2^{225}=\left(2^3\right)^{75}=8^{75}< 9^{75}=\left(3^2\right)^{75}=3^{150}\)
b) \(2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\)
c) \(99^{20}=\left(99^2\right)^{10}=9801^{10}< 9999^{10}\)
Bài 3:
a) \(12^8.9^{12}=\left(4.3\right)^8.9^{12}=4^8.3^8.9^{12}=2^{16}.9^4.9^{12}=2^{16}.9^{16}=\left(2.9\right)^{16}=18^{16}\)
b) \(75^{20}=\left(75^2\right)^{10}=5625^{10}=\left(45.125\right)^{10}=45^{10}.125^{10}=45^{10}.5^{30}\)
a) 2x . 4 = 128
2x = 128 : 4
2x = 32
x = 32 : 2
x = 16
b)x . 17 = x
=> x = 0
Lời giải:
a.
\(3^{21}=3.3^{20}=3.9^{10}\)
\(2^{31}=2.2^{30}=2.(2^3)^{10}=2.8^{10}\)
Mà $3.9^{10}> 2.8^{10}$ nên $3^{21}> 2^{31}$
b.
$2^{300}=(2^3)^{100}=8^{100}$
$3^{200}=(3^2)^{100}=9^{100}$
Mà $8^{100}< 9^{100}$ nên $2^{300}< 3^{200}$
c.
$32^9=(2^5)^9=2^{45}$
$18^{13}> 16^{13}=(2^4)^{13}=2^{52}$
Mà $2^{45}< 2^{52}$ nên $32^9< 18^{13}$
a/
\(27^{81}=\left(3^3\right)^{81}=3^{241}\)
\(81^{27}=\left(3^4\right)^{27}=3^{108}\)
\(\Rightarrow27^{81}=3^{241}>3^{108}=81^{27}\)
b/
\(5^{60}=\left(5^3\right)^{20}=125^{20}\)
\(7^{40}=\left(7^2\right)^{20}=49^{20}\)
\(\Rightarrow5^{60}=125^{20}>49^{20}=7^{40}\)
c/
\(11^{102}=\left(11^2\right)^{51}=121^{51}>121^{50}>99^{50}\)
\(2^{2004}=\left(2^{668}\right)^3\)
\(5^{891}=\left(5^{297}\right)^3\)
mà \(2^{668}>5^{297}\)
nên \(2^{2004}>5^{891}\)
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