\(A=\left(\frac{1}{2^2}-1\right).\left(\frac{1}{3^2}-1\right)......\left(\frac{1}{10^2}-1\right)=\left(-\frac{3}{4}\right).\left(-\frac{8}{9}\right)......\left(-\frac{99}{100}\right)\)

\(A=\frac{\left(-3\right).\left(-8\right).....\left(-99\right)}{4.9........100}=\frac{\left(-1\right).3.\left(-2\right).4....\left(-9\right).11}{2.2.3.3.....10.10}=\frac{\left[\left(-1.-2.-3....-9\right).\left(3.4...11\right)\right]}{\left(2.3.....10\right).\left(2.3...10\right)}\)

\(A=\frac{\left(-1\right).11}{10.2}=\frac{-11}{20}< \frac{-10}{20}=\frac{-1}{2}\)

Suy ra \(A< -\frac{1}{2}\)

-A =( 1- 1/2 )(1 -1/3).....(1 -1/10)

    = 1/2 . 2/3 ..... 9/10

    = 1/10

-A = 1/10 nên A = -1/10

Vì 1/10 < 1/9 nên -1/10 > -1/9

Vậy A > -1/9

\(A=\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{10}-1\right)=-\frac{1}{2}.-\frac{2}{3}...-\frac{9}{10}\)

\(=\frac{-\left(1.2...9\right)}{2.3...10}=\frac{-1}{10}\)

A<-1/9

Nhận xét: \(\frac{1}{5}< \frac{1}{42};\frac{1}{9}< \frac{1}{42};\frac{1}{10}< \frac{1}{42};\frac{1}{40}< \frac{1}{42}\)

\(\Rightarrow S< \frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}+\frac{1}{42}\)

\(\Rightarrow S< \frac{5}{42}< \frac{21}{42}=\frac{1}{2}\)

Vậy S < 1/2

ta có: S=1159/2520 =>S<1/2

1/5+1/9+1/10+1/40+1/42=1159/2520

                      1159/2520=0.4599.....                  

                                1/2=0.5

Mà:0.5>0.4599

Nên:1/5+1/9+1/10+1/40+1/42>1/2

\(A=\frac{3^{10}+1}{3^9+1}=\frac{3^{10}+3-2}{3^9+1}=\frac{3\left(3^9+1\right)-2}{3^9+1}=3-\frac{2}{3^9+1}\)

\(B=\frac{3^9+1}{3^8+1}=\frac{3^9+3-2}{3^8+1}=\frac{3\left(3^8+1\right)-2}{3^8+1}=3-\frac{2}{3^8+1}\)

Có \(3^9+1>3^8+1\)

\(\Rightarrow\frac{2}{3^9+1}< \frac{2}{3^8+1}\)

\(\Rightarrow3-\frac{2}{3^9+1}>3-\frac{2}{3^8+1}\)

\(\Rightarrow A>B\)