2)Ta có: \(2^{332}< 2^{333}=\left(2^3\right)^{111}=8^{111}\)

              \(3^{223}>3^{222}=\left(3^2\right)^{111}=9^{111}\)

Vì \(8^{111}< 9^{111}\) mà \(2^{332}< 8^{111},3^{223}>9^{111}\) nên suy ra \(2^{332}< 3^{223}\)

Vậy \(2^{332}< 3^{223}\)

1) \(A=\dfrac{10^{2013}+1}{10^{2014}+1}\Rightarrow10A=\dfrac{10^{2014}+10}{10^{2014}+1}=\dfrac{10^{2014}+1}{10^{2014}+1}+\dfrac{9}{10^{2014}+1}=1+\dfrac{9}{10^{2014}+1}\)

\(B=\dfrac{10^{2014}+1}{10^{2015}+1}\Rightarrow10B=\dfrac{10^{2015}+10}{10^{2015}+1}=\dfrac{10^{2015}+1}{10^{2015}+1}+\dfrac{9}{10^{2015}+1}=1+\dfrac{9}{10^{2015}+1}\)Vì: \(10^{2014}+1< 10^{2015}+1\Rightarrow\dfrac{9}{10^{2014}+1}>\dfrac{9}{10^{2015}+1}\Rightarrow1+\dfrac{9}{10^{2014}+1}>1+\dfrac{9}{10^{2015}+1}\)

Nên suy ra \(10A>10B\Rightarrow A>B\)

\(10\cdot\dfrac{10^3+5}{10^4+5}=\dfrac{10^4+5+45}{10^4+5}=1+\dfrac{45}{10^4+5}\)

\(10\cdot\dfrac{10^2+5}{10^3+5}=\dfrac{10^3+5+45}{10^3+5}=1+\dfrac{45}{10^3+5}\)

mà \(\dfrac{45}{10^4+5}< \dfrac{45}{10^3+5}\)

nên \(\dfrac{10^3+5}{10^4+5}< \dfrac{10^2+5}{10^3+5}\)

\(A\)\(=\dfrac{3^{18}+2}{3^{19}+10}\)

\(17A=\dfrac{3^{19}+6}{3^{19}+10}=1-\dfrac{4}{3^{19}+10}\)

\(B=\dfrac{3^{17}+2}{3^{18}+10}\)

\(17B=\dfrac{3^{18}+6}{3^{18}+10}=1-\dfrac{4}{3^{18}+10}\)

Vì \(\dfrac{4}{3^{19}+10}< \dfrac{4}{3^{18}+10}\)

⇒\(1-\dfrac{4}{3^{19}+10}\) \(>\) \(1-\dfrac{4}{3^{18}+10}\)

⇒\(A>B\)