a) Ta có
$4^{336} = 4^{3.112} = (4^3)^{112} = 64^{112}$
$3^{448} = 3^{4.112} = (3^4)^{112} = 81^{112}$
Do 64 < 81 nên $64^{112} < 81^{112}$ hay $4^{336} < 3^{448}$.
b) Ta có
$A = 1 + 8 + 8^2 + \cdots + 8^{150}$
$8A = 8 + 8^2 + 8^3 + \cdots + 8^{150} + 8^{151}$
Vậy
$8A - A = (8 + 8^2 + 8^3 + \cdots + 8^{150} + 8^{151}) - (1 + 8 + 8^2 + \cdots + 8^{150}$
$<-> 7A = 8^{151}-1$
$<-> A = \dfrac{8^{151}-1}{7}$
Vậy $A = B$.