\(A=\frac{1000^9+2}{1000^9-1}=\frac{1000^9-1+3}{1000^9-1}=\frac{1000^9-1}{1000^9-1}+\frac{3}{1000^9-1}=1+\frac{3}{1000^9-1}\)

\(B=\frac{1000^9+1}{1000^9-2}=\frac{1000^9-2+3}{1000^9-2}=\frac{1000^9-2}{1000^9-2}+\frac{3}{1000^9-2}=1+\frac{3}{1000^9-2}\)

Vì \(1000^9-1>1000^9-2\Rightarrow\frac{3}{1000^9-1}< \frac{3}{1000^9-2}\Rightarrow1+\frac{3}{1000^9-1}< 1+\frac{3}{1000^9-2}\Rightarrow A< B\)

Vậy A < B

Đặt a=1000^2012 thì \(A=\frac{a+2}{a-1}\)   ;   \(B=\frac{a}{a-3}\)

Xét    \(A-B=\frac{a+2}{a-1}-\frac{a}{a-3}=\frac{\left(a+2\right)\left(a-3\right)-a\left(a-1\right)}{\left(a-1\right)\left(a-3\right)}\)

                      \(=\frac{a^2-a-6-a^2+a}{\left(a-1\right)\left(a-3\right)}=\frac{-6}{\left(a-1\right)\left(a-3\right)}\)

Do \(a>1;a>3\)  nên \(\left(a-1\right)\left(a-3\right)>0\Leftrightarrow A-B< 0\)

 Do đó \(A>B\)

Ta có: \(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdot\ldots\cdot\left(1-\frac{1}{200^2}\right)\)

\(=\left(1-\frac12\right)\left(1-\frac13\right)\cdot\ldots\cdot\left(1-\frac{1}{200}\right)\left(1+\frac12\right)\left(1+\frac13\right)\cdot\ldots\cdot\left(1+\frac{1}{200}\right)\)

\(=\frac12\cdot\frac23\cdot\ldots\cdot\frac{199}{200}\cdot\frac32\cdot\frac43\cdot\ldots\cdot\frac{201}{200}=\frac{1}{200}\cdot\frac{201}{2}=\frac{201}{400}>\frac{200}{400}=\frac12\)

Cho A = 1/3² + 1/3³ + 1/3⁴ + ... + 1/3⁹

=>3A=1/3+1/3²+1/3³+...+1/3⁸

=>3A-A=1/3+1/3²+1/3³+...+1/3⁸-1/3²-1/3³-1/3⁴-...-1/3⁹

=>2A=1/3-1/3⁹

=>\(A=\frac{\frac{1}{3}-\frac{1}{3^9}}{2}\)<1

Vậy A<1