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\(a,16^{19}=\left(2^4\right)^{19}=2^{76}\\ 8^{25}=\left(2^3\right)^{25}=2^{75}\)
Vì \(2^{76}>2^{75}=>16^{19}>8^{25}\)
b,\(3^{500}=\left(3^5\right)^{100}=243^{100}\)
Vì \(243^{100}>5^{100}=>3^{500}>5^{100}\)
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\(3^{99}=\left(3^3\right)^{33}=27^{33}>27^{21}>11^{21}\\ 16^x< 128^4\\ \Rightarrow\left(2^4\right)^x< \left(2^7\right)^4\\ \Rightarrow2^{4x}< 2^{28}\Rightarrow4x< 28\Rightarrow x< 7\)
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\(1,\\ 16^x< 128^4\Rightarrow\left(2^4\right)^x< \left(2^6\right)^4\Rightarrow2^{4x}< 2^{24}\\ \Rightarrow4x=24\Rightarrow x=6\\ 2,\\ 3^{99}=\left(3^3\right)^{33}=27^{33}>27^{21}>11^{21}\)
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a: Ta có: \(3^{2x+1}< 27\)
\(\Leftrightarrow2x+1< 3\)
\(\Leftrightarrow x< 1\)
hay x=0
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a: 99^20=9801^10<9999^10
b: 3^500=243^100
5^300=125^300
=>3^500>5^300
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Giải:
a) Gọi dãy đó là A, ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\)
\(2A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\)
\(2A-A=\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2013}}\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2014}}\right)\)
\(A=\dfrac{1}{2}-\dfrac{1}{2^{2014}}\)
Vì \(\dfrac{1}{2}< 1;\dfrac{1}{2^{2014}}< 1\) nên \(\dfrac{1}{2}-\dfrac{1}{2^{2014}}< 1\)
\(\Rightarrow A< 1\)
b) \(A=\dfrac{10^{11}-1}{10^{12}-1}\) và \(B=\dfrac{10^{10}+1}{10^{11}+1}\)
Ta có:
\(A=\dfrac{10^{11}-1}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-10}{10^{12}-1}\)
\(10A=\dfrac{10^{12}-1+9}{10^{12}-1}\)
\(10A=1+\dfrac{9}{10^{12}-1}\)
Tương tự:
\(B=\dfrac{10^{10}+1}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+10}{10^{11}+1}\)
\(10B=\dfrac{10^{11}+1+9}{10^{11}+1}\)
\(10B=1+\dfrac{9}{10^{11}+1}\)
Vì \(\dfrac{9}{10^{12}-1}< \dfrac{9}{10^{11}+1}\) nên \(10A< 10B\)
\(\Rightarrow A< B\)
Ta có : 1020 = (102)10 = 10010
Mà 9010 < 10010
Nên : 1020 > 9010
1020 và 9010
1020 = 102.10= ( 102)10
Ta thấy : 102 > 90 nên => ( 102)10 > 9010 hay 1020> 9010
Ps: mk ko chắc đâu!
^^ Học tốt nha!!!