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Xét B=\(\frac{2000+2001}{2001+2002}\)\(=\)\(\frac{2000}{2001+2002}\)\(+\)\(\frac{2001}{2001+2002}\)
Mà \(\frac{2000}{2001}>\frac{2000}{2001+2002}\); \(\frac{2001}{2002}>\frac{2001}{2001+2002}\) \(\Rightarrow\)\(\frac{2000}{2001}+\frac{2001}{2002}\)\(>\frac{2000+2001}{2001+2002}\)
Vậy \(A>B\)
Ta có \(\frac{2000}{2001}\approx1;\frac{2001}{2002}\approx1\Rightarrow A\approx2.\)\(\Rightarrow1< A< 2\)
\(2000+2001< 2001+2002\Rightarrow\frac{2000+2001}{2001+2002}< 1\)
Do đó A > B
A = 2000/2001 + 2001/2002 (1)
B = 2000+2001/ 2001+2002
=>\(B=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Vì\(\frac{2000}{2001+2002}< \frac{2000}{2001}\) (so sánh số cùng tử)
\(\frac{2001}{2001+2002}< \frac{2001}{2002}\) (2)
Từ (1)và (2)=> A>B
:
Ta có:
\(5^{2003}+5^{2002}+5^{2001}\)
\(=5^{2001}.\left(5^2+5+1\right)\)
\(=5^{2001}.31⋮31\)
\(\Rightarrow5^{2003}+5^{2002}+5^{2001}⋮31\left(đpcm\right)\)
a) Ta có :
\(5^{2003}+5^{2002}+5^{2001}\)
\(=5^{2001}\times5^2+5^{2001}\times5+5^{2001}\)
\(=5^{2001}\times\left(5^2+5+1\right)\)
\(=5^{2001}\times31\)
Vậy \(5^{2003}+5^{2002}+5^{2001}⋮31\)
b) Ta có :
\(4^{39}+4^{40}+4^{41}\)
\(=4^{39}+4^{39}\times4+4^{39}\times4^2\)
\(=4^{39}\times\left(1+4+4^2\right)\)
\(=4^{39}\times21\)
Vậy \(4^{39}+4^{40}+4^{41}⋮21\)
_Chúc bạn học tốt_
2002A= 2002 + \(2002^2+2002^3+2002^4+.....+2002^{100}\)
2002A - A= \(\left(2002+2002^2+2002^3+2002^4+....+2002^{100}\right)-\left(1+2002+2002^2+.....+2002^{99}\right)\)
2001A= \(2002^{100}-1\)
Vì \(2002^{100}\) > \(2002^{100}-1\) nên B > 2001A