So sánh.

- Hướng dẫn giải

a, Ta có ${10}^{30}={\left({10}^3\right)}^{10}={1000}^{10}$

$2^{100}={\left(2^{10}\right)}^{10}={\left(1024\right)}^{10}$

Vì 1000<1024 nên ${1000}^{10}$ < ${1024}^{10}$

Vậy ${10}^{30}$ < $2^{100}$

b, Ta có: ${333}^{444}={\left({333}^4\right)}^{111}={\left({\left(3.111\right)}^4\right)}^{111}$ = ${\left(81.{111}^4\right)}^{111}$

${444}^{333}={\left({444}^3\right)}^{111}={\left({\left(4.111\right)}^3\right)}^{111}$ = ${\left(64.{111}^3\right)}^{111}$

Vì 81 > 64 và ${111}^4>{111}^3$ nên ${\left(81.{111}^4\right)}^{111}$>${\left(64.{111}^3\right)}^{111}$

Vậy ${333}^{444}$>${444}^{333}$

c, Ta có: ${21}^5={\left(3.7\right)}^{15}=3^{15}.7^{15}$

${27}^5.{49}^8={\left(3^3\right)}^5.{\left(7^2\right)}^8=3^{15}.7^{16}$

Vì $7^{15}<7^{16}$ nên $3^{15}.7^{15}<3^{15}.7^{16}$

Vậy ${21}^5$ < ${27}^5.{49}^8$

d, Ta có: $3^{2n}={\left(3^2\right)}^n=9^n$

$2^{3n}={\left(2^3\right)}^n=8^n$

Vì 8 < 9 nên $8^n<9^n\left(n\in N^{*}\right)$

Vậy $3^{2n}$ > $2^{3n}$

e, Ta có: 2017.2018 = (2018–1).(2018+1) = 2018.2018+2018.1–1.2018–1.1

= ${2018}^2-1$

Vì ${2018}^2-1$<${2018}^2$ nên 2017.2018<${2018}^2$

f, Ta có: ${\left(100-99\right)}^{2000}=1^{2000}=1$

${\left(100+99\right)}^0={199}^0=1$

Vậy ${\left(100-99\right)}^{2000}$ = ${\left(100+99\right)}^0$

g, Ta có: ${2009}^{10}+{2009}^9={2009}^9.\left(2009+1\right)$

= $2010.{2009}^9$

${2010}^{10}=2010.{2010}^9$

Vì ${2009}^9<{2010}^9$ nên $2010.{2009}^9$ < $2010.{2010}^9$

Vậy ${2009}^{10}+{2009}^9$ < ${2010}^{10}$