C1:
\(A=\dfrac{10^{50}+2}{10^{50}-1}=\dfrac{10^{50}-1}{10^{50}-1}+\dfrac{3}{10^{50}-1}=1+\dfrac{3}{10^{50}-1}\\ B=\dfrac{10^{50}}{10^{50}-3}=\dfrac{10^{50}-3}{10^{50}-3}+\dfrac{3}{10^{50}-3}=1+\dfrac{3}{10^{50}-3}\\ \text{Vì }10^{50}-3< 10^{50}-1\Rightarrow\dfrac{3}{10^{50}-3}>\dfrac{3}{10^{50}-1}\Rightarrow1+\dfrac{3}{10^{50}-3}>1+\dfrac{3}{10^{50}-1}\Leftrightarrow B>A\)
Vậy \(B>A\)
C2: Áp dụng \(\dfrac{a}{b}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\left(n>0\right)\)
Dễ thấy
\(B=\dfrac{10^{50}}{10^{50}-3}>1\\ \Rightarrow B=\dfrac{10^{50}}{10^{50}-3}>\dfrac{10^{50}+2}{10^{50}-3+2}=\dfrac{10^{50}+2}{10^{50}-1}=A\)
Vậy \(B>A\)