Bài 2:
\(10M=\dfrac{10^{12}+10}{10^{12}+1}=1+\dfrac{9}{10^{12}+1}\)
\(10N=\dfrac{10^{11}+10}{10^{11}+1}=1+\dfrac{9}{10^{11}+1}\)
Ta có: \(10^{12}+1>10^{11}+1\)
=>\(\dfrac{9}{10^{12}+1}< \dfrac{9}{10^{11}+1}\)
=>\(\dfrac{9}{10^{12}+1}+1< \dfrac{9}{10^{11}+1}+1\)
=>10M<10N
=>M<N
Bài 1:
\(A=\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{900}\right)\)
\(=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\cdot...\cdot\left(1-\dfrac{1}{30}\right)\cdot\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot...\cdot\left(1+\dfrac{1}{30}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{29}{30}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{31}{30}\)
\(=\dfrac{1}{30}\cdot\dfrac{31}{2}=\dfrac{31}{60}\)
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