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a: \(\dfrac{121212}{212121}=\dfrac{4}{7}\)
\(\dfrac{12121212}{21212121}=\dfrac{4}{7}\)
Do đó: \(\dfrac{121212}{212121}=\dfrac{12121212}{21212121}\)
b: \(2+\dfrac{1}{-50}=\dfrac{99}{50}=1.98>0>-\dfrac{82}{4}\)
Vì \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < 1
Nên \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}\)
Ta có: \(\frac{2005^{2005}+1+2004}{2005^{2006}+1+2004}=\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}=\frac{2005^{2004}+1}{2005^{2005}+1}\)
Nên: \(\frac{2005^{2005}+1}{2005^{2006}+1}\) < \(\frac{2005^{2004}+1}{2005^{2005}+1}\)
=> A < B
\(A=\frac{3^{100}+1}{3^{99}+1}=\frac{\left(3^{99}+1\right)\times3-2}{3^{99}+1}=3-\frac{2}{3^{99}+1}\)
\(B=\frac{3^{99}+1}{3^{98}+1}=\frac{\left(3^{98}+1\right)\times3-2}{3^{98}+1}=3-\frac{2}{3^{98}+1}\)
Do 398 + 1 < 399 + 1
=> \(\frac{2}{3^{98}+1}>\frac{2}{3^{99}+1}\)
=> A > B
Có: 1030 = 103.10 = (103)10 = 100010
2100 = 210.10 = (210)10 = 102410
Vì 1000 < 1024 => 100010 < 102410 => 1030 < 2100
Ta có : \(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Vì :\(1000^{10}< 1024^{10}\)
\(\Rightarrow10^{30}< 2^{100}\)
hay :\(A< B\)
Ta có:
\(2^{80}< 2^{81}\)
Lại có:
\(2^{80}=\left(2^{10}\right)^8=1024^8\)
\(3^{24}=\left(3^3\right)^8=27^8\)
Ta thấy:
\(1024^8< 27^8\Rightarrow2^{80}< 3^{24}\)
Mà: \(2^{80}< 2^{81}\Rightarrow2^{81}>3^{24}\)
Vậy: \(2^{81}>3^{24}\)