Question

[Nguyễn Thùy Linh]

So sánh A và B biét

A=\(\frac{19^{30}+5}{10^{31}+5}\)và B=\(\frac{19^{31}+5}{19^{32}+5}\)

A= \(\frac{2^{18}-3}{2^{20}-3}\)và B = \(\frac{2^{20}-3}{2^{22}-3}\)

A = \(\frac{1+5+5^2+.......+5^9}{1+5+5^2+.....+5^8}\) B = \(\frac{1+3+3^2+.....+3^9}{1+3+3^2+.......+3^8}\)

Answer 1

[Nhật Hạ]

a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)

b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)

\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)

Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)

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