Question

Cho \(A=\dfrac{10^8+2}{10^8-1}\)và \(B=\dfrac{10^8}{10^8-3}\). So sánh A và B ?

Answer 1

Ta có :

\(A=\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=\dfrac{10^8-1}{10^8-1}+\dfrac{3}{10^8-1}=1+\dfrac{3}{10^8-1}\)

\(B=\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=\dfrac{10^8-3}{10^8-3}+\dfrac{3}{10^8-3}=1+\dfrac{3}{10^8-3}\)

Vì \(1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\Rightarrow A< B\)

~ Học tốt ~

Answer 2

\(A=\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=\dfrac{10^8-1}{10^8-1}+\dfrac{3}{10^8-1}=1+\dfrac{3}{10^8-1}\)

\(B=\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=\dfrac{10^8-3}{10^8-3}+\dfrac{3}{10^8-3}=1+\dfrac{3}{10^8-3}\)

Vì \(\dfrac{3}{10^8-1}< \dfrac{3}{10^8-3}\)

\(\Rightarrow1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\)

\(\Rightarrow\dfrac{10^8+2}{10^8-1}< \dfrac{10^8}{10^8-3}\)

Vậy A < B.