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a, Ta có: \(\left(\dfrac{1}{2}\right)^{300}=\left[\left(\dfrac{1}{2}\right)^3\right]^{100}=\left(\dfrac{1}{8}\right)^{100}\)
\(\left(\dfrac{1}{3}\right)^{200}=\left[\left(\dfrac{1}{3}\right)^2\right]^{100}=\left(\dfrac{1}{9}\right)^{100}\)
=> \(\left(\dfrac{1}{8}\right)^{100}>\left(\dfrac{1}{9}\right)^{100}\)=> \(\left(\dfrac{1}{2}\right)^{300}>\left(\dfrac{1}{3}\right)^{200}\)
b, Ta có: \(\left(\dfrac{1}{3}\right)^{75}=\left[\left(\dfrac{1}{3}\right)^3\right]^{25}=\left(\dfrac{1}{27}\right)^{25}\)
\(\left(\dfrac{1}{5}\right)^{50}=\left[\left(\dfrac{1}{5}\right)^2\right]^{25}\)\(=\left(\dfrac{1}{25}\right)^{25}\)
Do \(\left(\dfrac{1}{27}\right)^{25}< \left(\dfrac{1}{25}\right)^{25}=>\left(\dfrac{1}{3}\right)^{75}< \left(\dfrac{1}{5}\right)^{50}\)
Kiểm tra lại bài nhé, học tốt!!
Ta có :
\(\frac{-1}{2}^{300}=\left[\left(-\frac{1}{2}\right)^3\right]^{100}=\left(-\frac{1}{8}\right)^{100}\)
\(\frac{-1}{3}^{200}=\left[\left(-\frac{1}{3}\right)^2\right]^{100}=\frac{1}{9}^{100}\)
vì \(\left(-\frac{1}{8}\right)^{100}=\frac{1}{8}^{100}\)mà 8100 < 9100 nên \(\frac{1}{8}^{100}>\frac{1}{9}^{100}\)hay \(\left(-\frac{1}{8}\right)^{100}>\left(\frac{1}{9}\right)^{100}\)
Vậy \(\left(-\frac{1}{2}\right)^{300}>\left(-\frac{1}{3}\right)^{200}\)
\(\left(\frac{-1}{2}\right)^{300}=\left[\left(\frac{-1}{2}\right)^3\right]^{100}=\left(\frac{-1}{8}\right)^{100}\)
\(\left(\frac{-1}{3}\right)^{200}=\left[\left(\frac{-1}{3}\right)^2\right]^{100}=\left(\frac{1}{9}\right)^{100}\)
vì \(\left(\frac{-1}{8}\right)^{100}< \left(\frac{1}{9}\right)^{100}\)nên \(\left(\frac{-1}{2}\right)^{300}< \left(\frac{-1}{3}\right)^{200}\)
So sánh :
a) 3300 + 4300 và 3.24100
b) \(\frac{2^{23}+1}{2^{24}+1}\) và \(\frac{2^{24}+1}{2^{25}+1}\)
\(1,\\ a,2^x=16=2^4\Rightarrow x=4\\ b,3^{x+1}=9^x=3^{2x}\\ \Rightarrow x+1=2x\Rightarrow x=1\\ c,2^{3x+2}=4^{x+5}=2^{2\left(x+5\right)}\\ \Rightarrow3x+2=2x+10\Rightarrow x=8\\ d,3^{2x-1}=243=3^5\\ \Rightarrow2x-1=5\Rightarrow x=3\\ 2,\\ a,2^{225}=8^{75}< 9^{75}=3^{150}\\ b,2^{91}=\left(2^{13}\right)^7=8192^7>3125^7=\left(5^5\right)^7=5^{35}\\ c,99^{20}=\left(99^2\right)^{10}< \left(99\cdot101\right)^{10}=9999^{10}\\ 3,\\ a,12^8\cdot9^{12}=2^{16}\cdot3^8\cdot3^{24}=2^{16}\cdot3^{32}=\left(2\cdot3^2\right)^{16}=18^{16}\\ b,75^{20}=\left(3\cdot5^2\right)^{20}=3^{20}\cdot5^{40}=\left(3^{20}\cdot5^{10}\right)\cdot5^{30}=\left(3^2\cdot5\right)^{10}\cdot5^{30}=45^{10}\cdot5^{30}\)
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=1+2+2^2+2^3+....+2^9
2A=2+2^2+2^3+....+2^10
2A-A=2^10-1
A=2^10-1/2
B=5.2^8=(2^2+1).2^8=2^10+2^8
=>B>A
2A = 2(1 + 2 + 22 + .... + 29 )
= 2 + 22 + 23 + ..... + 210
2A - A = (2 + 22 + 23 + ..... + 210) - (1 + 2 + 22 + .... + 29 )
A = 210 - 1
B = 5.28 = (22 + 1).28 = 210 + 28
210 - 1 < 210 + 28
=> A < B
Ta có : 333^444=(3.111)^444=3^444.111^444
444^333=(4.111)^333=4^333.111^333
Ta lại có : 3^444=(3^4)^111=81^111
4^333=(4^3)^111=64^111
vì 3^444>4^333
mặt khác 111^333<111^444
suy ra 4^333.111^333<3^444.111^444
vậy 333^444>444^333