\(A=123456789.123456789\)

\(B=123456788.123456790\)

Ta có: B=123456788*123456790

          <=>B=(123456789-1)(123456789+1)

                   =123456789^2-1<123456789
Hay B<A

123456789.123456789

123456789.123456789

Ta có: B=123456788*123456790

<=> B=(123456789-1)(132456789+1)

          =123456789^2-1<123456789

Hay B<A

Không

1) A = 2003.2005 = 2003.2004 + 2003

    B = 2004² = 2004.2003 + 2004

=> A < B

2) A = 123456787.123456789 = 123456787.123456788 + 123456787

    B = 123456788² = 123456788.123456787 + 123456788

=> A < B

=1 nha

=1 kb nha

123456789 - 123456788 = 1

Đúng 100%

Đúng 100%

Đúng 100%

Ta có

\(A=\frac{2011}{123456789}+\frac{2011}{987654321}+\frac{1}{987654321}\)

\(B=\frac{2011}{123456789}+\frac{1}{123456789}+\frac{2011}{987654321}\)

Mặt khác

\(\frac{1}{987654321}< \frac{1}{123456789}\)

\(\Rightarrow\frac{2011}{123456789}+\frac{2011}{987654321}+\frac{1}{987654321}< \frac{2011}{123456789}+\frac{1}{123456789}+\frac{2011}{987654321}\)

=> A<B

So sánh A và B biết :

A = a^a^9 = a^9.a

B = 9^9^a = 9^9.a

Đây là hai lũy thừa cùng số mũ 

mà a = 123456789 > 9 suy ra A > B  nhé

A > B nhá ^_^"

123456789 (có 9 chữ số) < 10^9 (10 chữ số) < 81^9 = 9^18 
=> 123456789^9 < 9^(18*9) < 9^(123456789) 

Ta co: \(M=\frac{2013}{123456789}+\frac{2014}{987654321}=\frac{2013}{123456789}+\frac{2013}{987654321}+\frac{1}{987654321}\)

\(N=\frac{2013}{123456789}+\frac{1}{123456789}+\frac{2013}{987654321}\)

ma \(\frac{1}{987654321}< \frac{1}{123456789}\) nen \(M< N\)

\(M=\frac{2013}{123456789}+\frac{2014}{987654321}\)

\(N=\frac{2014}{123456789}+\frac{2013}{987654321}\)

\(M=\frac{2014}{987654321}-\frac{1}{987654321}\)

\(N=\frac{2014}{123456789}-\frac{1}{123456789}\)

Ta thấy \(\frac{1}{123456789}>\frac{1}{987654321}\)

\(\Rightarrow M< N\)