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+ \(\frac{2000}{2001}=\frac{2001-1}{2001}=1-\frac{1}{2001}\)
+ \(\frac{2001}{2002}=\frac{2002-1}{2002}=1-\frac{1}{2002}\)
+ \(\frac{1}{2001}>\frac{1}{2002}\Rightarrow1-\frac{1}{2001}
\(1-\frac{2000}{2001}=\frac{1}{2001}\)
\(1-\frac{2001}{2002}=\frac{1}{2002}\)
Vì \(\frac{1}{2001}>\frac{1}{2002}\) nên \(\frac{2000}{2001}
Ta có: 2000/2001 = 1 - 1/2001
2001/2002 = 1 - 1/2002
mà 1/2001 > 1/2002
--> 1 - 1/2001 < 1 - 1/2002
--> 2000/2001 < 2001/2002
\(\frac{2000}{2001}=1-\frac{1}{2001}\)
\(\frac{2001}{2002}=1-\frac{1}{2002}\)
\(2001< 2002\Rightarrow\frac{1}{2001}>\frac{1}{2001}\)
\(\Rightarrow1-\frac{1}{2001}< 1-\frac{1}{2002}\)
\(\Rightarrow\frac{2000}{2001}< \frac{2001}{2002}\)
ta có:2000/2001=1-1/2001
2001/2002=1-1/2002
mà 2001<2002
suy ra 1/2001>1/2002
suy ra 1-1/2001<1-1/2002
vậy 2000/2001<2001/2002
phần bù đến 1 của 2000/2001 là 1- 2000/2001=1/2001
phần bù đến 1 của 2001/2002 là 1-2001/2002=1/2002
Vì 1/2001>1/2002 nên 2000/2001<2001/2002
13/27 và 7/15
\(\frac{13}{27}\) = 1:\(\frac{27}{13}\)= 1: \(\frac{26+1}{13}\) = 1: ( 2+\(\frac{1}{13}\))
\(\frac{7}{15}\)= 1:\(\frac{15}{7}\)= 1: \(\frac{14+1}{7}\)= 1: ( 2+ \(\frac{1}{7}\))
ta có \(\frac{1}{13}\)< \(\frac{1}{7}\)=> 2+\(\frac{1}{13}\)< 2+ \(\frac{1}{7}\) => 1: ( 2+\(\frac{1}{13}\)) > 1: ( 2+ \(\frac{1}{7}\))
vậy \(\frac{13}{27}\)>\(\frac{7}{15}\)- 2000/2001 và 2001/2002
\(\frac{2000}{2001}\)= \(\frac{2001-1}{2001}\)= 1 - \(\frac{1}{2001}\)
\(\frac{2001}{2002}\)= \(\frac{2002-1}{2002}\)= 1 - \(\frac{1}{2002}\)
ta có \(\frac{1}{2001}\)> \(\frac{1}{2002}\) => 1 - \(\frac{1}{2001}\) < 1 - \(\frac{1}{2002}\)
vậy \(\frac{2000}{2001}\)< \(\frac{2001}{2002}\)
2001/2000=1+1/2000
2002/2001=1+1/2001
Mà 1/2000>1/2001
=>1+1/2000>1+1/2001
hay 2001/2000>2002/2001
\(a,\dfrac{199}{200}=1-\dfrac{1}{200};\dfrac{200}{201}=1-\dfrac{1}{201}\\ Vì:\dfrac{1}{200}>\dfrac{1}{201}\\ \Rightarrow1-\dfrac{1}{200}< 1-\dfrac{1}{201}\\ Vậy:\dfrac{199}{200}< \dfrac{200}{201}\\ b,\dfrac{2001}{2002}=1-\dfrac{1}{2002};\dfrac{2002}{2003}=1-\dfrac{1}{2003}\\ Vì:\dfrac{1}{2002}>\dfrac{1}{2003}\Rightarrow1-\dfrac{1}{2002}< 1-\dfrac{1}{2003}\\ Vậy:\dfrac{2001}{2002}< \dfrac{2002}{2003}\)
\(c,\dfrac{2021}{2020}=1+\dfrac{1}{2020};\dfrac{2020}{2019}=1+\dfrac{1}{2019}\\ Vì:\dfrac{1}{2020}< \dfrac{1}{2019}\\ Nên:1+\dfrac{1}{2020}< 1+\dfrac{1}{2019}\\ Vậy:\dfrac{2021}{2020}< \dfrac{2020}{2019}\\ d,\dfrac{199}{198}=1+\dfrac{1}{198};\dfrac{200}{199}=1+\dfrac{1}{199}\\ Vì:\dfrac{1}{198}>\dfrac{1}{199}\\ Nên:1+\dfrac{1}{198}>1+\dfrac{1}{199}\\ Vậy:\dfrac{199}{198}>\dfrac{200}{199}\)
So sánh giữa 2000/2001 và 2001/2002
1 - 2000/2001 = 1/2001
1- 2001/2002 = 1/2002
Vì 1/2001 > 1/2002 nên 2000/2001 > 2001/2002
ta có:
1-\(\frac{2000}{2001}\)=\(\frac{2001}{2001}\)-\(\frac{2000}{2001}\)=\(\frac{1}{2002}\)
và 1- \(\frac{2001}{2002}\)=\(\frac{2002}{2002}\)-\(\frac{2001}{2002}\)=\(\frac{1}{2002}\)
vì \(\frac{1}{2001}\)> \(\frac{1}{2002}\)nên\(\frac{2000}{2001}\)<\(\frac{2001}{2002}\)
a) 2000/ 2002 > 2001/2002
b) 3/5 < 4/9
c) 16/60 < 31/90
cái này mà ko quy đồng thì mới là chuyện lạ có thật!!!