Ta thấy :  ${99}^2<99.101=9999\Rightarrow {\left({99}^2\right)}^{10}<{9999}^{10}\Rightarrow {99}^{20}<{9999}^{10}$

ta có :    ${11}^{1979}<{11}^{1980}={\left({11}^3\right)}^{660}={1331}^{660}$            (1)                                           

                    ${37}^{1320}={\left({37}^2\right)}^{660}={1369}^{660}$      (2)

                Từ   (1) và (2)   suy ra :  ${11}^{1979}<{37}^{1320}$ 

Ta có :    ${10}^{10}=2^{10}{.5}^{10}={2.2}^9{.5}^{10}$                                          (*)

                     ${48.50}^5=\left({3.2}^4\right).\left(2^5{.5}^{10}\right)={3.2}^9{.5}^{10}$                   (**)

                Từ (*) và (**)   $\Rightarrow {10}^{10}<{48.50}^5$   

Có :  ${1990}^{10}+{1990}^9={1990}^9.\left(1990+1\right)={1991.1990}^9$ 

              ${1991}^{10}={1991.1991}^9$ 

              Vì   ${1990}^9<{1991}^9$nên    ${1990}^{10}+{1990}^9<{1991}^{10}$ 

Ta có :      $2^{63}={128}^9$ 

                $5^{27}={125}^9$             

                $\Rightarrow 2^{63}>5^{27}$                                   (1)

    Lại có:   $$$2^{63}={512}^7$ 

                     $5^{28}={625}^7$                   

                $\Rightarrow 2^{63}<5^{28}$                                                        (2)

Từ (1) và (2)   $5^{27}<2^{63}<5^{28}$

Ta thấy :   ${107}^{50}<{108}^{50}={\left(4.27\right)}^{50}=2^{100}{.3}^{150}$                   (1)

                      ${73}^{75}>{72}^{75}={\left(8.9\right)}^{75}=2^{225}{.3}^{150}$                   (2)

                Từ (1) và (2)   $\Rightarrow {107}^{50}<2^{100}{.3}^{150}<2^{225}{.3}^{150}<{73}^{75}$

$2^{91}>2^{90}={32}^{18}$

         $5^{35}<5^{36}={25}^{18}$

        $\Rightarrow 2^{91}>{32}^{18}>{25}^{18}>5^{35}$

Vậy      $2^{91}>5^{35}$ 

Ta có   $A={27}^5={\left(3^3\right)}^5=3^{15}$

               $B={\left(3^5\right)}^3=3^{15}$ 

b)  $A=2^{300}=2^{3.100}=8^{100}$ 

      $B=3^{200}=3^{2.100}=9^{100}$

Vì $8<9$  nên $8^{100}<9^{100}$

$\Rightarrow A<B$

${199}^{20}<{200}^{20}={\left(2^3{.5}^2\right)}^{20}=2^{60}{.5}^{40}$

      ${2003}^{15}>{2000}^{15}={\left({2.10}^3\right)}^{15}={\left(2^4{.5}^3\right)}^{15}=2^{60}{.5}^{45}$                      

  Vậy  ${2003}^{15}>{199}^{20}$

${72}^{45}-{72}^{44}={72}^{44}.\left(72-1\right)={72}^{44}.71$

 ${72}^{44}-{72}^{43}={72}^{43}.\left(72-1\right)={72}^{43}.71$

Vậy ${72}^{45}-{72}^{44}>{72}^{44}-{72}^{43}$

Ta có:               $9^5={\left(3^2\right)}^5=3^{10}$

                        ${27}^3={\left(3^3\right)}^3=3^9$        

Ta có:   $3^{200}={\left(3^2\right)}^{100}=9^{100}$

                           $2^{300}={\left(2^3\right)}^{100}=8^{100}$

Ta có:  $3^{500}={\left(3^5\right)}^{100}={243}^{100}$

                  $7^{300}={\left(7^3\right)}^{100}={343}^{100}$

d) Ta có:  $8^5={\left(2^3\right)}^5=2^{15}$

               $2^{15}={2.2}^{14}<{3.2}^{14}={3.4}^7$

Ta có:  ${202}^{303}={\left({202}^3\right)}^{101}$;  ${303}^{202}={\left({303}^2\right)}^{101}$

Ta so sánh       ${202}^3$ và ${303}^2$

                         ${202}^3=2^3{\mathrm{.101.101}}^2$   

                         ${303}^2=3^2{.101}^2$

$2A=1.2+2.2+2^2.2+\mathrm{...}+2^4.2$

Vậy  $A=B$

$3C-C=\left(3^2+3^3+3^4+\mathrm{...}+3^{101}\right)-\left(3+3^2+3^3+\mathrm{...}+3^{100}\right)$

 $2C=3^2+3^3+3^4+\mathrm{...}+3^{101}-3-3^2-3^3-\mathrm{...}-3^{100}$

 $2C=3^{101}-3$

 $C=\frac{3^{101}-3}{2}$

Vậy  $C=D$