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Ta có: \(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19.\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19.\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Nên \(19A< 19B\Rightarrow A< B\)
Nhầm: Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow A>B\)
19A= \(\frac{19^{31}+95}{19^{31}+5}\) = \(\frac{19^{31}+95}{19^{31}}\)+\(\frac{19^{31}+95}{5}\)=95+ \(\frac{19^{31}+95}{5}\)
19B=\(\frac{19^{32}+95}{19^{32}+5}\)=\(\frac{19^{32}+95}{19^{32}}\)+\(\frac{19^{32}+95}{5}\)=95+\(\frac{19^{32}+95}{5}\)
vì \(\frac{19^{31}+95}{5}\)<\(\frac{19^{32}+95}{5}\)=> 19A<19B => A<B
Ta có:
\(A=\frac{19^{30}+5}{19^{31}+5}\Rightarrow19A=\frac{19\left(19^{30}+5\right)}{19^{31}+5}=\frac{19^{31}+95}{19^{31}+5}=\frac{19^{31}+5+90}{19^{31}+5}=1+\frac{90}{19^{31}+5}\)
\(B=\frac{19^{31}+5}{19^{32}+5}\Rightarrow19B=\frac{19\left(19^{31}+5\right)}{19^{32}+5}=\frac{19^{32}+95}{19^{32}+5}=\frac{19^{32}+5+90}{19^{32}+5}=1+\frac{90}{19^{32}+5}\)
Vì \(\frac{90}{19^{31}+5}>\frac{90}{19^{32}+5}\Rightarrow1+\frac{90}{19^{31}+5}>1+\frac{90}{19^{32}+5}\Rightarrow19A>19B\Rightarrow A>B\)
\(B=\frac{19^{31}+5}{19^{32}+5}<\frac{19^{31}+5+14}{19^{32}+5+14}=\frac{19^{31}+19}{19^{32}+19}=\frac{19\left(19^{30}+1\right)}{19\left(19^{31}+1\right)}=\frac{19^{30}+1}{19^{31}+1}<\frac{19^{30}+5}{19^{31}+5}=A\)
Vậy A>B
k mình bạn nhé TvT
C = 1930+5/1931+5
=>19C = 1931+95/1931+5 = 1+ [90/1931+5]
D = 1931+5/1932+5
=>19D = 1932+95/1932+5 = 1 + [90/1932+5]
ma 90/1931+5 > 90/1932+5
=>19C > 19D
=>C > D