ta có 50^40>50^39

        50^39>50^38=)1/50^39<1/50^38

=)50^40+1/50^39>50^39+1/50^38

=)50^40+1/50^39+1>50^39+1/50^38+1

=)A>B

Ta có A = \(\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}+\dfrac{1}{110}+\dfrac{1}{132}\)

= \(\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}+\dfrac{1}{10\cdot11}+\dfrac{1}{11\cdot12}\)

= \(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}\)

= \(\dfrac{1}{6}-\dfrac{1}{12}=\dfrac{1}{12}\) 

B = \(\dfrac{\dfrac{2}{29}-\dfrac{2}{39}+\dfrac{2}{49}}{\dfrac{23}{29}-\dfrac{23}{39}+\dfrac{23}{49}}=\dfrac{2\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}{23\left(\dfrac{1}{29}-\dfrac{1}{39}+\dfrac{1}{49}\right)}=\dfrac{2}{23}\) 

Lại có \(\dfrac{2}{23}>\dfrac{2}{24}=\dfrac{1}{12}\) hay A < B

Vậy A < B

Cảm ơn bạn nhé!

a) \(9^3\cdot3^2\)

\(=\left(3^2\right)^3\cdot3^2\)

\(=3^6\cdot3^2\)

\(=3^8\)

b) \(x^7\cdot x:x^4\)

\(=x^8:x^4\)

\(=x^4\)

c) \(7\cdot3^9+3^{10}+51\cdot3^8\)

\(=3^8\cdot\left(7\cdot3+3^2+51\right)\)

\(=3^8\cdot81\)

\(=3^8\cdot3^4\)

\(=3^{12}\)

d) \(5^{15}\cdot125^3\cdot625\)

\(=5^{15}\cdot5^9\cdot5^4\)

\(=5^{28}\)

A = 37³⁸+5/37³⁹+5 < 37³⁸+5+32 / 37³⁹+5+32

                            = 37³⁸+37 / 37³⁹⁺³⁷

                                  = 37(37³⁷+1) / 37(37³⁸+1)

                            = 37³⁷ + 1 / 37³⁸+1   = B

=> A < B nha!

Ai k mk mk k lại !!

a) ta có:  \(1-\frac{2012}{2013}=\frac{1}{2013}\)

                 \(1-\frac{2013}{2014}=\frac{1}{2014}\)

mà \(\frac{1}{2013}>\frac{1}{2014}\) nên   \(\frac{2013}{2014}>\frac{2012}{2013}\)

sao giống lớp 4 thế ta