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\(\frac{311}{256}\)và \(\frac{199}{203}\)
Vì \(\frac{311}{256}>1;1>\frac{199}{203}\)nên \(\frac{311}{256}>\frac{199}{203}\)
Học tốt #
a)
\(1-\frac{1998}{1999}=\frac{1}{1999}\)
\(1-\frac{1999}{2000}=\frac{1}{2000}\)
Vì \(\frac{1}{1999}>\frac{1}{2000}\)nên \(\frac{1998}{1999}< \frac{1999}{2000}\)
b) Ta có :
\(\frac{1999}{2001}< 1\)
\(\frac{12}{11}>1\)
Nên \(\frac{1999}{2001}< \frac{12}{11}\)
c)
\(1-\frac{13}{27}=\frac{14}{27}\)
\(1-\frac{27}{41}=\frac{14}{41}\)
Vì \(\frac{14}{27}>\frac{14}{41}\)nên \(\frac{13}{27}< \frac{27}{41}\)
d)
Ta có phân số trung gian là \(\frac{23}{45}\).
Ta có : \(\frac{23}{47}< \frac{23}{45}\) ; \(\frac{24}{45}>\frac{23}{45}\)
Nên \(\frac{23}{47}< \frac{24}{45}\)
A. 13/5 = 2 + 3/5 ; 15/7 = 2 + 1/7
3/5 > 1/7 => 13/5 > 15/7
B. 2012/1999 = 1 + 13/1999 ; 1999/1986 = 1 + 13/1986
13/1999 < 13/1986 => 2012/1999 < 1999/1986
C. 33/17 = 1 + 16/17
37/18 = 2 + 1/18
=> 33/17 < 37/18 => 17/33 > 18/37
D. 405/203 = 1 + 202/203
203/101 = 2 + 1/101
=> 405/203 < 203/101 => 203/405 > 101/203
`#3107.101107`
`a)`
Ta có:
\(\dfrac{2727}{3131}=\dfrac{2727\div27}{3131\div31}=\dfrac{27}{31}\)
Vì \(\dfrac{27}{31}=\dfrac{27}{31}\)
\(\Rightarrow\dfrac{27}{31}=\dfrac{2727}{3131}\)
`b)`
Ta có:
\(\dfrac{11}{31}=1-\dfrac{20}{31}=1-\dfrac{200}{310}\)
\(\dfrac{111}{311}=1-\dfrac{200}{311}\)
Vì \(\dfrac{200}{310}>\dfrac{200}{311}\)
\(\Rightarrow1-\dfrac{200}{310}< 1-\dfrac{200}{311}\)
\(\Rightarrow\dfrac{11}{31}< \dfrac{111}{311}.\)
\(a,\dfrac{199}{200}=1-\dfrac{1}{200};\dfrac{200}{201}=1-\dfrac{1}{201}\\ Vì:\dfrac{1}{200}>\dfrac{1}{201}\\ \Rightarrow1-\dfrac{1}{200}< 1-\dfrac{1}{201}\\ Vậy:\dfrac{199}{200}< \dfrac{200}{201}\\ b,\dfrac{2001}{2002}=1-\dfrac{1}{2002};\dfrac{2002}{2003}=1-\dfrac{1}{2003}\\ Vì:\dfrac{1}{2002}>\dfrac{1}{2003}\Rightarrow1-\dfrac{1}{2002}< 1-\dfrac{1}{2003}\\ Vậy:\dfrac{2001}{2002}< \dfrac{2002}{2003}\)
\(c,\dfrac{2021}{2020}=1+\dfrac{1}{2020};\dfrac{2020}{2019}=1+\dfrac{1}{2019}\\ Vì:\dfrac{1}{2020}< \dfrac{1}{2019}\\ Nên:1+\dfrac{1}{2020}< 1+\dfrac{1}{2019}\\ Vậy:\dfrac{2021}{2020}< \dfrac{2020}{2019}\\ d,\dfrac{199}{198}=1+\dfrac{1}{198};\dfrac{200}{199}=1+\dfrac{1}{199}\\ Vì:\dfrac{1}{198}>\dfrac{1}{199}\\ Nên:1+\dfrac{1}{198}>1+\dfrac{1}{199}\\ Vậy:\dfrac{199}{198}>\dfrac{200}{199}\)
\(a,\dfrac{2727}{3131}=\dfrac{2727:101}{3131:101}=\dfrac{27}{31}\\ Vậy:\dfrac{27}{31}=\dfrac{2727}{3131}\)
a) Ta có \(\dfrac{23}{27}>\dfrac{23}{29};\dfrac{23}{29}>\dfrac{22}{29}\)
Vậy \(\dfrac{23}{27}>\dfrac{22}{29}\)
b) Ta có \(\dfrac{15}{25}=1-\dfrac{2}{5}\)
\(\dfrac{25}{49}=1-\dfrac{24}{49}\)
Vì \(\dfrac{2}{5}=\dfrac{24}{60}< \dfrac{24}{49}\)
Vậy \(\dfrac{15}{25}>\dfrac{25}{49}\)
A/49/211<13/1999
B/311/256>199/203
C/26/27<96/27
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