a) 999 < 1000

    3000 > 2999

    8972 = 8972

    500 + 5 < 5005

b) 9999 > 9998

    9998 = 9990 + 8

    2009 < 2010

    7351 > 7153

1999 - 2000 + 2999 - 3000 + 3999 - 4000 + 4999 - 5000 + 5999 - 1000 

=(1999-2000)+(2999-3000)+(3999-4000)+(4999-5000)+(5999-1000)

=1+(-1)+(-1)+(-1)+4999

=1x4+4999

=4+4999

=4999-4

=4995

tk nha

=4995

k cho mik nhe

cac ban chinh bay cach giai ra nhe . day la tinh nhanh ma . 

1999-2000+2999-3000+3999-4000+4999-5000+5999-1000

=(1999-2000)+(2999-3000)+(3999-4000)+(4999-5000)+(5999-1000)

=-1+(-1)+(-1)+(-1)+4999

=-1x4+4999

=-4+4999

=4999-4

=4995

1999 - 2000 + 2999 - 3000 + 3999-4000+ 4999-5000+5999-1000

= ( 2000-1999) + ( 3000-2999) + ( 4000-3999) + ( 5999-5000) -1000

=  1 + 1 + 1 + 1 -1000

=4 -1000

= 1000-4

=996

1999-2000+2999-3000+3999-4000+4999-5000+5999-1000=4995

k cho mk nha năn nỉ đó

Câu 1:

B = \(\frac{2999}{1}+\frac{2998}{2}+\frac{2997}{3}+...+\frac{1}{2999}\)

= \(\frac{3000-1}{1}+\frac{3000-2}{2}+\frac{3000-3}{3}+...+\frac{3000-2999}{2999}\)

= \(\left(\frac{3000}{1}+\frac{3000}{2}+\frac{3000}{3}+...+\frac{3000}{2999}\right)-\left(\frac{1}{1}+\frac{2}{2}+\frac{3}{3}+...+\frac{2999}{2999}\right)\)

= \(3000+3000.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)-2999\)

= \(3000\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2999}\right)+\frac{3000}{3000}\)

= \(3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)\)

\(\Rightarrow\frac{A}{B}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}\right)}=\frac{1}{3000}\)

các bn ơi 

giúp mk đi mà

+.+

C=2013/1*2014/2*2015/3*...*3012/1000

C=2013*2014*2015*...*3012/1*2*3*...*1000

D=1001/1*1002/2*1003/3*...*3012/2012

D=1001*1002*...*3012/1*2*...*2012

Suy ra C/D=2013*2014*2015*...3012*1*2*...*2012/1*2*3*...*1000*1001*1002*...*3012

( Nhân đảo ngược)

Vậy C/D=1

Tổng của a và n là 1001

Vậy trung bình cộng của 3 số abc là : 

\(\left(1001+1000+2999\right):6=\frac{2500}{3}\)

\(3999+2999+2+3999+2999+2\\ =\left(3999+2999+2\right)+\left(3999+2999+2\right)\\ =2\cdot\left(3999+2999+2\right)\\ =2\cdot7000\\ =14000\)

\(3999+2999+2+3999+2999+2\)

= \(\left(3999+3999\right)+\left(2999+2999\right)+\left(2+2\right)\)

= \(7998+5998+4\)

= \(13996+4\)

= \(14000\).

Chúc bạn học tốt!

3999+2999+2+3999+2999+2

=(3999+3999)+(2999+2999)+(2+2)

=7998+5998+4

=13996+4=14000

Thế nhé

a)\(3^{1989}=3^{1988}.3=\left(3^4\right)^{497}.3=\left(...1\right).3=\left(...3\right)\)

b)\(2^{2999}+3^{2999}=2^{4.749}.2^3+3^{4.749}.3^3=\left(...6\right).8+\left(...1\right).27\)

\(=\left(...8\right)+\left(...7\right)\)

\(=\left(...5\right)\)

Xét mẫu :

\(\frac{2999}{1}+\frac{2998}{2}+.....+\frac{1}{2999}\)

=\(\left(1+\frac{2998}{2}\right)+\left(1+\frac{2997}{3}\right)+....+\left(1+\frac{1}{2999}\right)+1\)

=\(\frac{3000}{2}+\frac{3000}{3}+.....+\frac{3000}{2999}+\frac{3000}{3000}\)

=\(3000\left(\frac{1}{2}+\frac{1}{3}+....+\frac{1}{3000}\right)\)

Thay vào ta có:

\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{3000}}{3000\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{3000}\right)}\)

=\(\frac{1}{3000}\)