Câu hỏi của Baophuong Hoang

Question

so sánh10^8+2/10^8-1 và 10^8/10^8-3

Nguyễn Thanh Hằng · Accepted Answer

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 $1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\Rightarrow A< B$Câu trả lời: 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 $1+\dfrac{3}{10^8-1}< 1+\dfrac{3}{10^8-3}\Rightarrow A< B$