\(\frac{2010.125+1000}{126.2010-1010}=\frac{10\left(201.125+100\right)}{10\left(201.126-101\right)}=\frac{201.125-101+201}{2011.126-101}\)

\(=\frac{201.126-101}{201.126-101}=1\)

=2010 x 125 + 2010x 126 + 1000+1010

=2010 x 125 + 2010 x 126 + 2010

=2010 x ( 125 + 126 +1 )

=2010 x 252

=506520

\(\frac{2010.125+1000}{126.2010-1010}=\frac{2010.125+1000}{2010.125+2010-1010}=\frac{2010.125+1000}{2010.125+1000}=\frac{1}{1}=1\)

A = 2006 x 125 + 1000 / 126 x 2005 - 888

A = (2005 + 1) x 125 + 1000 / (125 + 1) x 2005 - 888

A = 2005 x 125 + 125 + 1000 / 125 x 2005 + 2005 - 888

A = 2005 x 125 + 1125 / 125 x 2005 + 1117

A = 1125 / 1117

8 x 2010 x 125 - 2009 x 437 - 2009 x 563

= ( 8 x 125 ) x 2010 - 2009 x ( 437 + 563 )

= 1000 x 2010 - 2009 x 1000

= 1000 x ( 2010 - 2009 )

= 1000 x 1

= 1000

=))

8 x 2010 x 125 - 2009 x 437 - 2009 x 563 = 1000

8 x 2010 x 125 - 2009 x 437 - 2009 x 563

(125 x 8) x 2010 - (2009 x 437 + 2009 x 563)

= 1000 x 2010 - [2009 x (437 + 563)]

= 1000 x 2010 - 2009 x 1000

= 1000 x (2010 - 2009)

= 1000 x 1

= 1000

\(\frac{2010\cdot2011+1000}{2012\cdot2010-1010}\)

= \(\frac{2010\cdot2011+1000}{\left(2011+1\right)\cdot2010-1010}\)

= \(\frac{2010\cdot2011+1000}{2011\cdot2010+2010-1010}\)

= \(\frac{2010\cdot2011+1000}{2011\cdot2010+1000}\)

= 1

\(\frac{2010.2011+1000}{2012.2010-1010}\)

\(=\frac{2010.2011+2010-1010}{2012.2010-1010}\)

\(=\frac{2010.\left(2011+1\right)-1010}{2012.2010-1010}\)

\(=\frac{2010.2012-1010}{2012.2010-1010}\)

\(=1\)

\(\frac{2010\cdot2011+1000}{2012\cdot2010-1010}\)= \(\frac{2010\cdot2011+1000}{2011\cdot2010+2010-1010}\)= \(\frac{2010\cdot2011+1000}{2011\cdot2010+1000}\)= 1

         Bạn nguyen quang huy sai rồi!!!

Vì 1000/2009>1000/2009+2010 (1)

     1010/2010>1010/2009+2010  (2)

  Ta cộng theo vế (1) và (2) với nhau nên ta được:

    1000/2009+1010/2010>1000/2009+2010 +1010/2009+2010

=>1000/2009+1010/2010>1000+1010/2009+2010

Vậy A<B

Chắc chắn 100% luôn, không sai đâu!!!!!!!

bằng nhau vì 2 biểu thức giống nhau

Lời giải:

$(2012\times 2010+2010\times 2008)\times (1+\frac{1}{2}: 1\frac{1}{2}-1\frac{1}{3})$

$=2010\times (2012+2008)\times (1+\frac{1}{2}\times \frac{2}{3}-1\frac{1}{3})$

$=2010\times 4020\times (1+\frac{1}{3}-1\frac{1}{3})$

$=2010\times 4020\times 0=0$

=0 nha

\(A=\frac{2011\times2012}{2011+2012}+\frac{2009\times2010}{2009+2010}\)

\(A=\frac{2011\times2011}{2011+2012}+\frac{2011}{2011+2012}+\frac{2010\times2010}{2009+2010}-\frac{2010}{2009+2010}\)

\(A=\frac{2011\times2011}{2011+2012}+\frac{2010\times2010}{2009+2010}+\frac{2011}{2011+2012}-\frac{2010}{2009+2010}\)

\(A=B+\frac{2011}{2011+2012}-\frac{2010}{2009+2010}\)

\(A=B+\frac{2011}{4023}-\frac{2010}{4019}\)

Dễ thấy \(\frac{2011}{4023}-\frac{2010}{4019}< 0\)

\(\Rightarrow A< B\)

Bổ sung đề: So sánh A và B

Ta có:

A. \(2010^{1000}=\frac{1010^{1010}.2010^{1000}}{2010^{2010}}=\left(\frac{101}{201}\right)^{1010}\)

B. \(2010^{1000}=\frac{2010^{2010}.2010^{1000}}{3010^{3010}}=\left(\frac{201}{301}\right)^{3010}\)

Từ \(\frac{101}{201}>\frac{1}{2}>\frac{40401}{90601}=\left(\frac{201}{301}\right)^2\)và \(\frac{201}{301}< 1\)

  có: \(\left(\frac{101}{201}\right)^{1010}>\left(\frac{201}{301}\right)^{2.1010}=\left(\frac{201}{301}\right)^{2020}>\left(\frac{201}{301}\right)^{3010}\)

Suy ra \(A=\left(\frac{101}{201}\right)^{1010}.\frac{1}{2010^{1000}}>\left(\frac{201}{301}\right)^{3010}.\frac{1}{2010^{1000}}\) hay A > B