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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 \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)
c, Câu hỏi của truong nguyen kim
\(A=\frac{2^{18}-3}{2^{20}-3}=1-\frac{2^2}{2^{20}-3}\)
\(B=\frac{2^{20}-3}{2^{22}-3}=1-\frac{2^2}{2^{22}-3}\)
Vì \(\frac{2^2}{2^{20}-3}>\frac{2^2}{2^{22}-3}\) nên A < B
Ta có :
\(\dfrac{1}{A}=\dfrac{2^{20}-3}{2^{18}-3}=\dfrac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\dfrac{9}{2^{18}-3}\left(1\right)\)
\(\dfrac{1}{B}=\dfrac{2^{22}-3}{2^{20}-3}=\dfrac{2^2.\left(2^{20}-3\right)+9}{2^{20}-3}=4+\dfrac{9}{2^{20}-3}\left(2\right)\)
Từ (1) và (2) ta có \(\dfrac{1}{A}>\dfrac{1}{B}\Leftrightarrow A< B\)
Ta có :
\(A=\frac{2^{18}-3}{2^{20}-3}\) \(B=\frac{2^{20}-3}{2^{22}-3}\)
\(2^2A=\frac{2^{20}-12}{2^{20}-3}\) \(2^2B=\frac{2^{22}-12}{2^{22}-3}\)
\(4A=\frac{2^{20}-3-9}{2^{20}-3}\) \(4B=\frac{2^{22}-3-9}{2^{22}-3}\)
\(4A=1-\frac{9}{2^{20}-3}\) \(4B=1-\frac{9}{2^{22}-3}\)
Ta thấy 220 - 3 < 222 - 3 nên \(\frac{9}{2^{20}-3}>\frac{9}{2^{22}-3}\)hay \(1-\frac{9}{2^{20}-3}>1-\frac{9}{2^{22}-3}\)
=> 4A > 4B hay A > B
Vậy A > B
Ủng hộ mk nha !!! ^_^
A<B
tích giúp mình nha