mình lớp5  nhưng mình bt làm

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        \(A>B\)

A=1-2000/2001=2001/2001-2000/2001=1/2001

B=1-2000/2001=2001/2001-2000/2001=1/2001

Ta thấy 1/2001=1/2001 Nên 2000/2001=2000/2001

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_

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