So sanh :
A = \(\frac{2000}{2001}\) + \(\frac{2001}{2002}\)
B = \(\frac{2000+2001}{2001+2002}\)
Những câu hỏi liên quan
so sanh
\(A\frac{2000}{2001}+\frac{2001}{2002};B\frac{2000+2001}{2001+2002}\)
Ta có:\(B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}\)
Vì:\(\frac{2000}{2001}>\frac{2000}{2001+2002}\)
\(\frac{2001}{2002}>\frac{2001}{2001+2002}\)
\(\Rightarrow\left(\frac{2000}{2001}+\frac{2001}{2002}\right)>\left(\frac{2000}{2001-2002}-\frac{2001}{2001+2001}\right)\)
\(\Rightarrow A>B\)
Đúng 0
Bình luận (0)
so sanh A va B: \(A=\frac{2000}{2001}\) +\(\frac{2001}{2002}\)\(B=\frac{2000+2001}{2001+2002}\)
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 \(A>B\)
Đúng 0
Bình luận (0)
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
Đúng 0
Bình luận (0)
\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
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
Đúng 0
Bình luận (0)
SO SÁNH \(A=\frac{2000}{2001}+\frac{2001}{2002}v\text{à}B=\frac{2000+2001}{2001+2002}\)
Ta có:
B = \(\frac{2000}{2001+2002}\)+ \(\frac{2001}{2001+2002}\)
Vì \(\frac{2000}{2001}\)> \(\frac{2000}{2001+2002}\)
\(\frac{2001}{2002}\)> \(\frac{2001}{2001+2002}\)
=> \(\left(\frac{2000}{2001}+\frac{2001}{2002}\right)\)> \(\left(\frac{2000}{2001+2002}+\frac{2001}{2001+2001}\right)\)
=> A>B
Vậy A>B
Đúng 0
Bình luận (0)
so sánh : A= \(\frac{2000}{2001}+\frac{2001}{2002}\) B= \(\frac{2000+2001}{2001+2002}\)
Ta có:B= \(\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Vì \(\frac{2000}{2001}>\frac{2000}{2001+2002}\)và \(\frac{2001}{2002}>\frac{2001}{2001+2002}\)
Nên A>B
Đúng 0
Bình luận (0)
So sánh hai biểu thức A và B cho biết rằng:
\(A=\frac{2000}{2001}+\frac{2001}{2002}\) \(B=\frac{2000+2001}{2001+2002}\)
Ta có: B = \(\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}=\frac{2000}{4003}+\frac{2001}{4003}\)
Ta thấy : \(\frac{2000}{2001}>\frac{2000}{4003}\)(1)
\(\frac{2001}{2002}>\frac{2001}{4003}\) (2)
Từ (1) và (2) cộng vế với vế, ta được :
\(\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{4003}+\frac{2001}{4003}\)
hay \(A=\frac{2000}{2001}+\frac{2001}{2002}>B=\frac{2000+2001}{2001+2002}\)
Đúng 0
Bình luận (0)
So sanh
A = 2000/2001 +2001/2002
B = 2000+2001/2001+2002
Ta có:
\(A=\frac{2000}{2001}+\frac{2001}{2002}\) và \(B=\frac{2000+2001}{2001+2002}\)
\(\Rightarrow B=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
Ta Xét:
\(\frac{2000}{2001}>\frac{2000}{2001+2002}\)
\(\frac{2001}{2002}>\frac{2001}{2001+2002}\)
\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
\(\Rightarrow A>B\)
Đúng 0
Bình luận (0)
so sanh :
A=2000/2001+2001/2002
B=2000+2001/2001+2002
\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}
Đúng 0
Bình luận (0)
So Sánh 2 Biểu Thức:
\(A=\frac{2000}{2001}+\frac{2001}{2002}\)
\(B=\frac{2000+2001}{2001+2002}\)
B=2000/2001+2002 + 2001/2001+2002
Ta có:2000/2001 > 2000/2001+2002
2001/2002 > 2001/2001+2002
Vậy A >B
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
So sánh A và B :
A=\(\frac{2000}{2001}\)+\(\frac{2001}{2002}\)
B =\(\frac{2000+2001}{2001+2002}\)
giúp mik vs nhé mik cảm ơn
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