Question

so sánh

A=\(\frac{19^{20}+5}{19^{20}-8}\)

B=\(\frac{19^{20}+6}{19^{20}-7}\)

Answer 1

Ta có: \(A=\frac {19^{20}+5}{19^{20}-8}=\frac {19^{20}-8+13}{19^{20}-8}=1+\frac {13}{19^{20}-8}\)

           \(B=\frac {19^{20}+6}{19^{20}-7}=\frac {19^{20}-7+13}{19^{20}-7}=1+\frac {13}{19^{20}-7}\)

Vì \(19^{20}-8<19^{20}-7\) nên \(\frac {13}{19^{20}-8}>\frac {13}{19^{20}-7}\)

\(\Rightarrow\)\(1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\) Hay \(A>B\) 

Vậy A>B

Answer 2

ta có A = \(\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\) 

và B = \(\frac{19^{20}+6}{19^{20}-7}=\frac{19^{20}-7+13}{19^{20}-7}=1+\frac{13}{19^{20}-7}\)

vì \(\frac{13}{19^{20}-8}>\frac{13}{19^{20}-7}\)\(\Rightarrow1+\frac{13}{19^{20}-8}>1+\frac{13}{19^{20}-7}\)\(\Rightarrow A>B\)