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Cách 1
\(A=\frac{37^{20}}{37^{20}-6}=\frac{37^{20}-6+6}{37^{20}-6}=1+\frac{6}{37^{20}-6}\)
\(B=\frac{37^{20}+4}{37^{20}-2}=\frac{37^{20}-2+6}{37^{20}-2}=1+\frac{6}{37^{20}-2}\)
Vì \(\frac{6}{37^{20}-6}>\frac{6}{37^{20}+2}\Rightarrow1+\frac{6}{37^{20}-6}>1+\frac{6}{37^{20}+2}\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\frac{37^{20}}{37^{20}:37^6}=\frac{1}{\frac{1}{37^6}}=37^6\left(1\right)\)
\(B=\frac{37^{20}.37^4}{37^{20}:37^2}=\frac{37^4}{\frac{1}{37^2}}=37^6\left(2\right)\)
Từ (1)(2) => A = B
![](https://rs.olm.vn/images/avt/0.png?1311)
ta thấy:
\(A=\dfrac{37^{20}}{37^{20}-6}< \dfrac{37^{20}+4}{37^{20}-6}< \dfrac{37^{20}+4}{37^{20}-2}\)
\(\Rightarrow A< \dfrac{37^{20}+4}{37^{20}-2}\)
vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức : \(\frac{a}{b}< 1\Rightarrow\frac{a}{b}< \frac{a+n}{b+n}\)
Ta chứng minh được \(\frac{20}{39}>\frac{18}{41};\frac{18}{43}>\frac{14}{39};\frac{22}{27}>\frac{22}{29}\)
\(\Rightarrow\frac{20}{39}+\frac{22}{27}+\frac{18}{43}>\frac{14}{37}+\frac{22}{29}+\frac{18}{41}\)
\(\Rightarrow A>B\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có : 648 = (43)8 = 424
1612 = (42)12 = 424
Vì 424 = 424 => 648 = 1612