đề sai

1234

​k nhé

Đáp án cần chọn là: C

Vì trong tích có một thừa số bằng 0 nên M=0

a, 

= 2354 - 45 - 2354 

= -45

b, 

= 16 + 23 + 153 - 16 - 23

= 153

c, 

= -3752 - 29 + 3632 - 51

= ( 3632 - 3752 ) - ( 29 + 51 )

= -120 - 80

= -200

a, 

= 2354 - 45 - 2354 

= -45

b, 

= 16 + 23 + 153 - 16 - 23

= 153

c, 

= -3752 - 29 + 3632 - 51

= ( 3632 - 3752 ) - ( 29 + 51 )

= -120 - 80

= -200

a)2354-45-2345=2354-2354-45=-45. b)-2009-243+2009=-2009+2009-234+-243. c)16+23+153-16-23=16-16+23-23+153=153.d)134-167+45-134-45=134-134+45-45-167=-167.e)215+43-215-25=215-215+43-25=18.f)-312-327-28+27-312-28-327+27=340-300=40

a, (2354-45)-2354

=2354-45-2354

=(2354-2354)-45

=0-45

=-45

a: 43/52>26/52=1/2=60/120

b: 17/68=1/4<1/3=35/105<35/103

c: \(\dfrac{2018\cdot2019-1}{2018\cdot2019}=1-\dfrac{1}{2018\cdot2019}\)

\(\dfrac{2019\cdot2020-1}{2019\cdot2020}=1-\dfrac{1}{2019\cdot2020}\)

2018*2019<2019*2020

=>-1/2018*2019<-1/2019*2020

=>\(\dfrac{2018\cdot2019-1}{2018\cdot2019}< \dfrac{2019\cdot2020-1}{2019\cdot2020}\)

\(\dfrac{19}{19}\) = 1 < \(\dfrac{2005}{2004}\) vậy \(\dfrac{19}{19}\) < \(\dfrac{2005}{2004}\)

\(\dfrac{72}{73}\) = 1 - \(\dfrac{1}{73}\) 

\(\dfrac{98}{99}\) = 1 - \(\dfrac{1}{99}\)

Vì \(\dfrac{1}{73}\) > \(\dfrac{1}{99}\) nên \(\dfrac{72}{73}\) < \(\dfrac{98}{99}\) 

1) 19/19 < 2005/2004

2)72/73 > 98/99

a) ta có:  \(1-\frac{2012}{2013}=\frac{1}{2013}\)

                 \(1-\frac{2013}{2014}=\frac{1}{2014}\)

mà \(\frac{1}{2013}>\frac{1}{2014}\) nên   \(\frac{2013}{2014}>\frac{2012}{2013}\)

sao giống lớp 4 thế ta