so sánh biểu thức A và B biết rằng ;
A =2000/2001+2001/2001 B=2000+2001/2001+2002
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Ta có:
\(\frac{2000}{2001}\)> \(\frac{2000}{2001+2002}\)(1)
\(\frac{2001}{2002}\)> \(\frac{2001}{2001+2002}\)(2)
Cộng các bất đẳng thức (1) và ( 2) vế với nhau:
Vậy \(\frac{2000}{2001}\)+ \(\frac{2001}{2002}\)> \(\frac{2000+2001}{2001+2002}\)hay A > B.
B=2000/2001+2002 + 2001/2001+2002
Ta có:
2000/2001 > 2000/2001+2002
2001/2002 > 2001/2001+2002
Vậy A >B
\(B=\frac{2000}{2001}+2002+\frac{2001}{2001}+2002\)
Ta có: \(\frac{2000}{2001}>\frac{2000}{2001}+2002\)
\(\frac{2001}{2002}>\frac{2001}{2001}+2002\)
Vậy A>B
Ta có:\(\frac{2018}{2019}\)<1\(\Rightarrow\)\(\frac{2018}{2019}\)>\(\frac{2018}{2019+2020}\)
\(\frac{2019}{2020}\)<1\(\Rightarrow\)\(\frac{2019}{2020}\)>\(\frac{2019}{2019+2020}\)
\(\Rightarrow\)\(\frac{2018}{2019}\)+\(\frac{2019}{2020}\)>\(\frac{2018}{2019+2020}\)+\(\frac{2019}{2019+2020}\)=\(\frac{2018+2019}{2019+2020}\)
\(\Rightarrow\)A>B
Vậy A>B
Ta có :\(A=\frac{2018}{2019}+\frac{2019}{2020}\)
\(B=\frac{2018+2019}{2019+2020}\)
\(B=\frac{2018}{2019+2020}+\frac{2019}{2019+2020}\)
Ta thấy :
\(\frac{2018}{2019}>\frac{2018}{2019+2020}\left(2019< 2019+2020\right)\)
\(\frac{2019}{2020}>\frac{2019}{2019+2020}\left(2020< 2019+2020\right)\)
\(\Rightarrow\frac{2018}{2019}+\frac{2019}{2020}>\frac{2018+2019}{2019+2020}\)
Vậy \(A>B\)
~ Thiên Mã ~
Ta có: B = \(\frac{2012+2013}{2013+2014}=\frac{2012}{2013+2014}+\frac{2013}{2013+2014}\)
Mà : \(\frac{2013}{2014}>\frac{2013}{2013+2014}\)và \(\frac{2012}{2013}>\frac{2012}{2013+2014}\)
=> A > B
k nhé
Ta có
B= 2000/2001+2002 + 2001/2001+2002. Mà 2000/2001+2002 < 2000/2001 và 2001/2001+2002 < 2001/2002. Nên 2000/2001+2002 + 2001/ 2001+2002 < 2000/2001 + 2001/2002. Hay 2000+2001/ 2001+2002 < 2000/2001 + 2001/2002 Suy ra B < A
A = 2,8 x 47,44 x 9,008
A = 28 x 0,1 x 474,4 x 0,1 x 9,008
A = 28 x 474,4 x 0,09008 = B
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 \(A=\frac{2000}{2001}+\frac{2001}{2002}>B=\frac{2000+2001}{2001+2002}\)
Đề sai chỗ 2001/2001 phải là 2001/2002
\(A=\dfrac{2000}{2001}+\dfrac{2001}{2002}>\dfrac{2000}{2002}+\dfrac{2001}{2002}=\dfrac{4001}{2002}>1\)
B=\(\dfrac{2000+2001}{2001+2002}=\dfrac{4001}{4003}< 1\)
=>A>B
A > B