Đáp án:

Giải thích các bước giải:

 a)
Ta có $99^{20} = (100 - 1)^{20} < 100^{20} = (10^2)^{20} = 10^{40}$
$9999^{10} > 1000^{10} = (10^3)^{10} = 10^{30}$
Vì $10^{40} > 10^{30}$
$\Rightarrow 99^{20} > 9999^{10}$

`b)`
Ta có $3^{21} = (3^7)^3 = 2187^3$, $2^{31} = (2^{10})^3 \cdot 2 = 1024^3 \cdot 2$
Vì $2187 > 1024$ nên $3^{21} > 2^{31}$

`c)`
$2^{30} + 3^{30} + 4^{30} < 3 \cdot 4^{30} = 3 \cdot (2^2)^{30} = 3 \cdot 2^{60}$
$3.24^{10} = (3 \cdot 24)^{10} = 72^{10} = (2^3 \cdot 3^2)^{10} = 2^{30} \cdot 3^{20}$
$\dfrac{2^{60}}{2^{30} \cdot 3^{20}} = \dfrac{2^{30}}{3^{20}}$
Vì $\dfrac{2^{30}}{3^{20}} < 1$ nên $2^{60} < 2^{30} \cdot 3^{20}$
$\Rightarrow 2^{30} + 3^{30} + 4^{30} < 3.24^{10}$