So sánh hai biểu thức A và B biết rằng
A = 2000/2001 + 2001/2002
B = 2000+2001/2001+2002
Lm hộ mk nha. Ths
không quy đồng mẫu số so sánh A=2000/2001+2001/2002
B=2000/2001+2001/2002
Ta có: \(A=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)và\(B=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)
Mà\(\dfrac{2000}{2001}+\dfrac{2001}{2002}=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)
Vậy A=B
Ta có: \(A=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)
\(B=\dfrac{2000}{2001}+\dfrac{2001}{2002}\)
Do đó: A=B
so sánh phân số :
\(\dfrac{2001}{2000}và\dfrac{2002}{2001}\)
So sánh 2000/2001+ 2001/2002+...+2015/2016 và 15
2000/2001<1
2001/2002<1
2002/2003<1
...
2015/2016<1
=>2000/2001+2001/2002+2002/2003+2003/2004+...+2015/2016<1+1+1+1+1+...+1=15
Vậy...
Ta có:
2000/ 2001 < 1
2001/2002 < 1
..................
2015/ 2016<1
=> 200/2001 + 2001/202+...+ 2015/2016 < 1 + 1+1 +1+...+1( 15 số hạng)
=> 200/2001 + 2001/202+...+ 2015/2016< 1 x 15 = 15
so sánh hai biểu thức A và B bk rằg:
A=2000/2001+2001/2002
B=2000+2001/2001+2002
\(\dfrac{2000}{2001}+\dfrac{1}{2001}=1\) mà \(\dfrac{1}{2001}< \dfrac{1}{2}\) nên \(\dfrac{2000}{2001}>\dfrac{1}{2}\)
\(\dfrac{2001}{2002}+_{ }\dfrac{1}{2002}=1\) mà \(\dfrac{1}{2002}< \dfrac{1}{2}\)nên \(\dfrac{2001}{2002}>\dfrac{1}{2}\)
- Hay A>1
- mà 2000+2001<2001+2002 nên B< 1
=> A>B
A=2000/2001 + 2001/2000 ; B=2000/2001+2001/2002
khong quy dong. So sanh A va B
Ta có: 2000/2001>1/2 ; 2001/2002>1/2
=>A=1/2+1/2=1=>A>1
B=2000+2001/2001+2002=4001/4003<1
A>1;B<1
=>A>B
Vậy A>B
$B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}$B=20002001+2002 +20012001−2002
Vì:
Ta có : \(B=\frac{2000}{2001+2002}+\frac{2001}{2001-2002}\)
Vì : \(\frac{2000}{2001}>\frac{2000}{2001+2002}\)
\(\frac{2001}{2002}>\frac{2001}{2001+2002}\)
\(\Rightarrow\left(\frac{2000}{2001}+\frac{2001}{2002}\right)>\left(\frac{2000}{2001-2002}-\frac{2001}{2001+2001}\right)\)
\(\Rightarrow A>B\)
So sánh A và B, biết: A= 2000/2001 + 2001/ 2002 và B= 2000 + 2001/ 2001 + 2002
ta có:\(A=\frac{2000}{2001}+\frac{2001}{2002}<\frac{2000}{2002}+\frac{2001}{2002}=\frac{2000+2001}{2002}<\frac{2000+2001}{2001+2002}=B\)
\(\Rightarrow A
ta có:\(B=\frac{2000+2001}{2001+2002}=\frac{2000}{2001+2002}+\frac{2001}{2001+2002}\)
vì \(\frac{2000}{2001}>\frac{2000}{2001+2002}và\frac{2001}{2002}>\frac{2001}{2001+2002}\)
\(\Rightarrow\frac{2000}{2001}+\frac{2001}{2002}>\frac{2000+2001}{2001+2002}\)
=>A>B
SO SÁNH HAI BIỂU THỨC A VÀ B2000/2001+2001/2002;B=2000+2001/2001+2002
So sánh:
A= 2000/2001 + 2001/2002 với. B=2000+2001/2001+2002
so sánh biểu thức A và B biết rằng ;
A =2000/2001+2001/2001 B=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
Ta có :
B = \(\dfrac{2000+2001}{2001+2002}\) \(=\dfrac{2000}{2001+2002}+\dfrac{2001}{2001+2002}\)
Vì \(\dfrac{2000}{2001}\) > \(\dfrac{2000}{2001+2002}\)
\(\dfrac{2001}{2002}\) \(>\dfrac{2001}{2001+2002}\)
Nên \(\dfrac{2000}{2001}+\dfrac{2001}{2002}\) \(>\dfrac{2000+2001}{2001+2002}\)
Vậy A > B