Ta có:
\(A=\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
\(B=\frac{19^{21}+6}{19^{21}-7}=\frac{19^{21}-7+13}{19^{21}-7}=1+\frac{13}{19^{21}-7}\)
Vì \(19^{20}-8< 19^{21}-7\Rightarrow\frac{13}{19^{20}-8}>\frac{13}{19^{21}-7}\)
\(\Rightarrow A>B\)
A =B NHA ! maivananh
ĐỀU = 1 CẢ .
Ta có : \(A=\frac{19^{20}+5}{19^{20}-8}=\frac{19^{20}-8+13}{19^{20}-8}=\frac{19^{20}-8}{19^{20}-8}+\frac{13}{19^{20}-8}=1+\frac{13}{19^{20}-8}\)
\(B=\frac{19^{21}+6}{19^{21}-7}=\frac{19^{21}-7+13}{19^{21}-7}=\frac{19^{21}-7}{19^{21}-1}+\frac{13}{19^{21}-7}=1+\frac{13}{19^{21}-7}\)
Mà : \(\frac{13}{19^{20}-8}< \frac{13}{19^{21}-7}\)
Nên : \(A< B\)
\(A=\frac{19^{20}-8+13}{19^{20}-8}=\frac{13}{19^{20}-8}\)+1
\(B=\frac{19^{21}-7+13}{19^{21}-7}=\frac{13}{19^{21}-7}+1\)
Suy ra A>B