Đáp án:

Giải thích các bước giải:

Bài `1:`

`a) 4^6 . 64^5 : 16^7` 

`=(2^2)^6 . (2^6)^5 : (2^4)^7`

`=2^12 . 2^30 : 2^28`

`=2^42 : 2^28`

`=2^14`

`b) 45^3 . 20^4 . 18^2 : 180^5`

`=(3^2 .5)^3 . (2^2 .5)^4 . (2.3^2)^2 : (2^2. 3^2. 5)^5`

`=3^6 . 5^3 . 2^8 . 5^4 . 2^2 . 3^4 : 2^10 : 3^10 : 5^5`

`=3^(6+4-10) . 5^(3+4-5) . 2^(8+2-10)`

`=3^0 . 5^2 . 2^0`

`=5^2`

`c) 3^4 . 4^6 . 9^5 : 6^12`

`=3^4 . (2^2)^6 . (3^2)^5 : (2.3)^12`

`=3^4 . 2^12 . 3^10 : 2^12 : 3^12`

`=3^(4+10-12) . 2^(12-12)`

`=3^2`

`d) 17^18 . 9^9`

`=17^18 . (3^2)^9`

`=17^18 . 3^18`

`=(17.3)^18`

`=51^18`

Bài `2:`

`a) 3^2=9; 2^3=8`

`=> 3^2>2^3`

`b) 3^50=(3^5)^10=243^10`

`7^30=(7^3)^10=343^10`

Mà `243^10<343^10`

`=> 3^50<7^30`

`c) 202^303=(202^3)^101`

303^202=(303^2)^101`

Mà `202^3>303^2`

`=> (202^3)^101> (303^2)^101`

`=> 202^303>303^202`

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
{4^6}{.64^5}:{16^7}\\
 = {\left( {{2^2}} \right)^6}.{\left( {{2^6}} \right)^5}:{\left( {{2^4}} \right)^7}\\
 = {2^{2.6}}{.2^{6.5}}:{2^{4.7}}\\
 = {2^{12}}{.2^{30}}:{2^{28}}\\
 = {2^{12 + 30 - 28}} = {2^{14}}\\
b,\\
{45^3}{.20^4}{.18^2}:{180^5}\\
 = {\left( {{{5.3}^2}} \right)^3}.{\left( {{2^2}.5} \right)^4}.{\left( {{3^2}.2} \right)^2}:{\left( {{2^2}{{.3}^2}.5} \right)^5}\\
 = {5^3}.{\left( {{3^2}} \right)^3}.{\left( {{2^2}} \right)^4}{.5^4}.{\left( {{3^2}} \right)^2}{.2^2}:\left[ {{{\left( {{2^2}} \right)}^5}.{{\left( {{3^2}} \right)}^5}{{.5}^5}} \right]\\
 = {5^3}{.3^{2.3}}{.2^{2.4}}{.5^4}{.3^{2.2}}{.2^2}:\left( {{2^{2.5}}{{.3}^{2.5}}{{.5}^5}} \right)\\
 = {5^3}{.3^6}{.2^8}{.5^4}{.3^4}{.2^2}:{2^{10}}:{3^{10}}:{5^5}\\
 = {5^{3 + 4 - 5}}{.3^{6 + 4 - 10}}{.2^{8 + 2 - 10}}\\
 = {5^2}{.3^0}{.2^0}\\
 = {5^2}\\
c,\\
{3^4}{.4^6}{.9^5}:{6^{12}}\\
 = {3^4}.{\left( {{2^2}} \right)^6}.{\left( {{3^2}} \right)^5}:{\left( {2.3} \right)^{12}}\\
 = {3^4}{.2^{2.6}}{.3^{2.5}}:\left( {{2^{12}}{{.3}^{12}}} \right)\\
 = {3^4}{.2^{12}}{.3^{10}}:{2^{12}}:{3^{12}}\\
 = {3^{4 + 10 - 12}}{.2^{12 - 12}}\\
 = {3^2}{.2^0} = {3^2}\\
d,\\
{17^{18}}{.9^9} = {17^{18}}.{\left( {{3^2}} \right)^9} = {17^{18}}{.3^{2.9}} = {17^{18}}{.3^{18}} = {\left( {17.3} \right)^{18}} = {51^{18}}\\
2,\\
a,\\
{3^2} = 9 > 8 = {2^3}\\
b,\\
{3^{50}} = {3^{5.10}} = {\left( {{3^5}} \right)^{10}} = {243^{10}} < {343^{10}} = {\left( {{7^3}} \right)^{10}} = {7^{3.10}} = {7^{30}}\\
c,\\
{202^{303}} = {202^{3.101}} = {\left( {{{202}^3}} \right)^{101}} > {\left( {{{303}^2}} \right)^{101}} = {303^{202}}
\end{array}\)