ta có : \(\frac{10^9+2}{10^9-1}=\frac{10^9}{10^9-3}\)
\(\Leftrightarrow\left(10^9+2\right)\left(10^9-3\right)=\left(10^9-1\right)10^9\)
\(\Leftrightarrow10^{18}-10^9.3+2.10^9-6=10^{18}-10^9\)
\(\Rightarrow10^{18}-10^9.3+2.10^9-6=10^{18}-\left(10^9.3-2.10^9+6\right)\)
\(=10^{18}-\left(10^9+6\right)\)
vì \(-10^9>-\left(10^9+6\right)\Rightarrow10^{18}-10^9>10^{18}-\left(10^9+6\right)\)
\(\Rightarrow A>B\)
Ta có: A=\(\frac{10^9+2}{10^9-1}=\frac{10^9-1+3}{10^9-1}=1+\frac{3}{10^9-1}\)
B=\(\frac{10^9}{10^9-3}=\frac{10^9-3+3}{10^9-3}=1+\frac{3}{10^9-3}\)
Mà \(\frac{3}{10^9-1}< \frac{3}{10^9-3}\Rightarrow1+\frac{3}{10^9-1}< 1+\frac{3}{10^9-3}\Rightarrow A< B\)
Vậy A<B
Ta có :
\(A=\frac{10^9+2}{10^9-1}=\frac{10^9-1+3}{10^9-1}=1+\frac{3}{10^9-1}\)
\(B=\frac{10^9}{10^9-3}=\frac{10^9-3+3}{10^9-3}=1+\frac{3}{10^9-3}\)
Vì \(10^9-1>10^9-3\) nên \(\frac{3}{10^9-1}< \frac{3}{10^9-3}\) \(\Rightarrow1+\frac{3}{10^9-1}< 1+\frac{3}{10^9-3}\)
Do đó : \(A< B\)