hello 123-145=

a) 3/4 < 1

115/14 > 1 

=> 3/4 < 115/14

b) 2000/1999 > 1

1998/1999 < 1

=> 2000/1999>1998/1999

c)-17/234 < 0 , 1/1995 > 0

=> -17/234 < 1/1995

d) 23/-15 < 0 ; -17/-49>0

=> 23/-15 < -17/-49

\(A=1999\times1995=\left(1997+2\right)\times\left(1997-2\right)=1997\times1997-2\times2=1997\times1997-4\\ ---\\ B=1994\times2000=\left(1997-3\right)\times\left(1997+3\right)=1997\times1997-3\times3=1997\times1997-9\\ Vì:1997\times1997=1997\times1997\\ Và:4< 9\\ Nên:1997\times1997-4>1997\times1997-9\\ Vậy:1999\times1995>1994\times2000\)

Sửa lại câu hỏi thành toán lớp mấy đó nhé!

\(A=1999\cdot1995=\left(2000-1\right)\cdot1995\)

\(A=2000\cdot1995-1995\)

\(B=1994\cdot2000=\left(1995-1\right)\cdot2000\)

\(B=1995\cdot2000-2000\)

Vì \(2000>1995\) nên \(2000\cdot1995-1995>2000\cdot1995-2000\) hay \(A>B\)

A = 1999 . 1995

= (1997 - 2)(1997 + 2)

= 1997² - 2²

= 1997² - 4

B = 1994 . 2000

= (1997 - 3)(1997 + 3)

= 1997² - 3²

= 1997² - 9

Do 4 < 9 nên 1997² - 4 > 1997² - 9

Vậy 1999 . 1995 > 1994 . 2000

\(B=\frac{2014x2016+2000}{2015x2015+1999}\)

\(B=\frac{2014x2015+(2014+2000)}{2014x2015+(2015+1999)}\)

\(B=\frac{2014x2015+4014}{2014x2015+4014}=1\)

=> B = 1

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 : \(\frac{2000}{2001}>\frac{2000}{2001+2002}\)

\(\frac{2001}{2002}>\frac{2001}{2001+2002}\)

\(\Rightarrow\left(\frac{2000}{2001}+\frac{2001}{2002}\right)>\left(\frac{2000}{2001-2002}-\frac{2001}{2001+2001}\right)\)

\(\Rightarrow A>B\)