Question

so sanh A va B:           \(A=\frac{2000}{2001}\) +\(\frac{2001}{2002}\)\(B=\frac{2000+2001}{2001+2002}\)

Answer 1

Ta có:

B=\(\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)

Do \(\frac{2000}{2001}>\frac{2000}{2001+2002};\frac{2001}{2002}>\frac{2001}{2001+2002}\)

\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)

\(\Rightarrow A>B\)

Vậy \(A>B\)

Answer 2

Ta có:$B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}$B=20002001+2002 +20012001−2002 

Vì:$\frac{2000}{2001}>\frac{2000}{2001+2002}$20002001 >20002001+2002 

$\frac{2001}{2002}>\frac{2001}{2001+2002}$20012002 >20012001+2002 

$\Rightarrow\left(\frac{2000}{2001}+\frac{2001}{2002}\right)>\left(\frac{2000}{2001-2002}-\frac{2001}{2001+2001}\right)$⇒(20002001 +20012002 )>(20002001−2002 −20012001+2001 )

$\Rightarrow A>B$⇒A>B

 

Answer 3

\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)

Mà \(\frac{2000}{2001}>\frac{2000}{2001+2002}\) và \(\frac{2001}{2002}>\frac{2001}{2001+2002}\)

nên \(\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\) hay A>B