Question

So sánh 

A= \(\frac{9^{2008}+1}{9^{2009}+1}\)và \(\frac{9^{2009}+1}{9^{2010}+1}\)

Answer 1

áp dụng \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có : \(B=\frac{9^{2009}+1}{9^{2010}+1}< 1\)

\(\Rightarrow B< \frac{9^{2009}+1+8}{9^{2010}+1+8}\)

\(\Rightarrow B< \frac{9.\left(9^{2008}+1\right)}{9.\left(9^{2009}+1\right)}=\frac{9^{2008}+1}{9^{2009}+1}\)

Vậy B < A

Answer 2

    B = 9²⁰⁰⁹ + 1/9²⁰¹⁰ + 1 < 1

=> B < 9²⁰⁰⁹ + 1 + 8 / 9²⁰¹⁰ + 1 + 8 = 9²⁰⁰⁹ + 9 / 9²⁰¹⁰ + 9 = 9 (9²⁰⁰⁸ + 1 ) / 9 ( 9²⁰⁰⁷ + 1) = A

=>B < A 

              #Hoq chắc _ Baccanngon

Answer 3

\(\frac{9^{2008}+1}{9^{2009}+1}=\frac{9^{2009}+9}{9^{2010}+9}>\frac{9^{2009}+1}{9^{2010}+1}\)