\(A=\dfrac{1995.1994-1}{1993.1995+1994}=\dfrac{1995\left(1993+1\right)-1}{1993.1995+1994}=\dfrac{1995.1993+1995-1}{1993.1995+1994}=\dfrac{1995.1993+1994}{1995.1993-1994}=1\)\(B=\dfrac{2004.2004+3006}{2005.2005-1003}=\dfrac{2004.2004+2004.1+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2005.2005-1003}=\dfrac{2004.2005+1002}{2004.2005+2005-1003}=\dfrac{2004.2005+1002}{2004.2005+1002}=1\)\(C=\dfrac{2010.2011-1}{2009.2011+2010}=\dfrac{2009.2011+2011-1}{2009.2011+2010}=\dfrac{2019.2011+2010}{2009.20011+2010}=1\)\(D=\dfrac{2014.2015-1}{2013.2015+2013}=\dfrac{2013.2015+2014-1}{2013.2015+2013}=\dfrac{2013.2015+2013}{2013.2015+2013}=1\)

Câu 1 nhầm đề nha bạn mình sửa:

\(\dfrac{1995.1994-1}{1993.1995+1994}\)

\(=\dfrac{1995.\left(1993+1\right)-1}{1993.1995+1994}\)

\(=\dfrac{1995.1993+1995-1}{1993.1995+1994}\)

\(=\dfrac{1993.1995+1994}{1993.1995+1994}\)

\(=1\)

Câu 2: \(\dfrac{2004.2004+3006}{2005.2005-1003}\)

\(=\dfrac{2004.2004+2004+1002}{\left(2004+1\right).\left(2004+1\right)-1003}\)

\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1-1003}\)

\(=\dfrac{2004.2004+2004+1002}{2004.2004+2004+1002}\)

\(=1\)

Câu 3:\(\dfrac{2010.2011-1}{2009.2011+2010}\)

\(=\dfrac{\left(2009+1\right).2011-1}{2009.2011+2010}\)

\(=\dfrac{2009.2011+2011-1}{2009.2011+2010}\)

\(=\dfrac{2009.2011+2010}{2009.2011+2010}\)

= 1

Câu 4:Nhầm để, sửa:

\(\dfrac{2014.2015-1}{2013.2015+2014}\)

\(=\dfrac{\left(2013+1\right).2015-1}{2013.2015+2014}\)

\(=\dfrac{2013.2015+2015-1}{2013.2015+2014}\)

\(=\dfrac{2013.2015+2014}{2013.2015+2014}\)

\(=1\)

Ta có:

2010 . 2011/2010 . 2011 + 1                                                                      2009 . 2010/2009 . 2010 + 1

= 1 - 1/2010 . 2011 + 1                                                                             = 1 - 1/2009 . 2010 + 1

Vì 2010 . 2011 + 1 > 2009 . 2010 + 1

=> 1/2010 . 2011 + 1 < 1/2009 . 2010 + 1

=> 1 - 1/2010 . 2011 + 1 > 1 - 1/2009 . 2010 + 1

=> 2010.2011/2010.2011+1 > 2009.2010/2009.2010+1

lớn hơn á bạn

\(\Rightarrow\frac{1}{2010\cdot2011+1}< \frac{1}{2009\cdot2010+1}\)

\(\Rightarrow-\frac{1}{2010\cdot2011+1}>-\frac{1}{2009\cdot2010+1}\)

\(\Rightarrow1-\frac{1}{2010\cdot2011+1}>1-\frac{1}{2009\cdot2010+1}\)

\(\Rightarrow\frac{2010\cdot2011+1-1}{2010\cdot2011+1}>\frac{2009\cdot2010+1-1}{2009\cdot2010+1}\)

\(\Rightarrow\frac{2010\cdot2011}{2010\cdot2011+1}>\frac{2009\cdot2010}{2009\cdot2010+1}\)(ĐPCM)

Biểu thức không rút gọn được. Bạn xem lại xem.

a=1.2 + 2.3 +3.4+ ...+ 2010.2011\

3a=1.2.3+2.3.3+3.4.3+......+2010.2011.3

3a=1.2.3+2.3.(4-1)+3.4.(5-1)+............+2010.2011.(2012-2009)

3a=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+....+2010.2011.2012-2009.2010.2011

3a=2010.2011.2012

a=2010.2011.2012:3

a=?

A=2010.2011.2012:3

Bài làm:

\(\frac{215}{300}=\frac{43}{60}\)

\(\frac{1200}{2010}=\frac{40}{67}\)

\(\frac{105}{420}=\frac{1}{4}\)

\(\frac{125}{420}=\frac{25}{84}\)

#Học_tốt

\(\dfrac{215}{300}\) = \(\dfrac{215:5}{300:5}\) = \(\dfrac{43}{60}\)

\(\dfrac{1200}{2010}\) = \(\dfrac{1200:30}{2010:30}\) = \(\dfrac{40}{67}\)

\(\dfrac{105}{420}\)  = \(\dfrac{105:105}{420:105}\) = \(\dfrac{1}{4}\)

\(\dfrac{125}{420}\) = \(\dfrac{125:5}{420:5}\) = \(\dfrac{25}{84}\)

Câu hỏi của Mun - Toán lớp 5 - Học toán với OnlineMath

Bạn vào đây xem

A = 2011.2010−12009.2011+2010 

 => \(\frac{2011.2009+2011-1}{2009.2011+2010}\) = \(\frac{2011.2009+2010}{2009.2011+2010}=1\) 

=> A = 1

t i c k nhé!! 354543564576856925253456456545756756887676976

\(A=\frac{2011.2010-1}{2009.2011+2010}\)

\(A=1\)

nha bạn                                      ^_^

Vì phân số cần tìm bằng phân số \(\frac{5}{7}\) nên phân số đó có dạng \(\frac{5a}{7a}\)

Vì tống của cả tử và mẫu của phân số đó là 4812 => 5a + 7a = 4812 

<=> 12a = 4812 => a = 4812 : 12 = 401

Vậy phân số \(\frac{5.401}{7.401}=\frac{2005}{2807}\)

Ta có:

\(\begin{array}{l}\frac{{24}}{{108}} = \frac{{24:12}}{{108:12}} = \frac{2}{9};\\\frac{{80}}{{32}} = \frac{{80:16}}{{32:16}} = \frac{5}{2}\end{array}\)