a) ${{202}^{303}}$ và ${{303}^{202}}$
Ta có: ${{202}^{303}}={{\left( {{\left( 2.101 \right)}^{3}} \right)}^{101}}={{\left( {{8.101}^{3}} \right)}^{101}}={{\left( {{808.101}^{2}} \right)}^{101}}$
${{303}^{202}}={{\left( {{\left( 3.101 \right)}^{2}} \right)}^{101}}={{\left( {{9.101}^{2}} \right)}^{101}}$
$0 < {{9.101}^{2}} < {{808.101}^{2}}\Rightarrow {{\left( {{9.101}^{2}} \right)}^{101}}{{303}^{202}}$
b) ${{\left( -32 \right)}^{9}}$ và ${{\left( -18 \right)}^{13}}$
Ta có: ${{\left( -32 \right)}^{9}}={{\left( -{{2}^{5}} \right)}^{9}}={{\left( -2 \right)}^{45}} > {{\left( -2 \right)}^{52}}={{\left( -{{2}^{4}} \right)}^{13}}={{\left( -16 \right)}^{13}} > {{\left( -18 \right)}^{13}}$
hay ${{\left( -32 \right)}^{9}} > {{\left( -18 \right)}^{13}}$
c) ${{5}^{21}}$ và ${{124}^{10}}$
Ta có: ${{5}^{21}}={{\left( {{5}^{2}} \right)}^{10}}.5={{25}^{10}}.5$
$\frac{124}{25} > 4\Rightarrow {{\left( \frac{124}{25} \right)}^{10}}>{{4}^{10}} > 5\Rightarrow {{124}^{10}} > {{25}^{10}}.5$
hay ${{5}^{21}} < {{124}^{10}}$
d) ${{\left( -5 \right)}^{30}}$ và ${{\left( -3 \right)}^{50}}$
Ta có: ${{\left( -5 \right)}^{30}}={{\left( -125 \right)}^{10}}$, ${{\left( -3 \right)}^{50}}={{\left( -243 \right)}^{10}}$
$0 < -243 < -125\Rightarrow {{\left( -125 \right)}^{10}} < {{\left( -243 \right)}^{10}}$ hay ${{\left( -5 \right)}^{30}} < {{\left( -3 \right)}^{50}}$
e) ${{\left( \frac{-1}{4} \right)}^{8}}$ và ${{\left( \frac{1}{8} \right)}^{50}}$
Ta có: ${{\left( \frac{-1}{4} \right)}^{8}}={{\left( \frac{1}{4} \right)}^{8}}={{\left( \frac{1}{{{2}^{2}}} \right)}^{8}}=\frac{1}{{{2}^{16}}}$
${{\left( \frac{1}{8} \right)}^{50}}={{\left( \frac{1}{{{2}^{3}}} \right)}^{50}}=\frac{1}{{{2}^{150}}}$
${{2}^{16}} < {{2}^{150}}\Rightarrow \frac{1}{{{2}^{16}}} > \frac{1}{{{2}^{150}}}$ hay ${{\left( \frac{-1}{4} \right)}^{8}} > {{\left( \frac{1}{8} \right)}^{150}}$
g) ${{\left( \frac{3}{8} \right)}^{5}}$ và ${{\left( \frac{5}{243} \right)}^{3}}$
Ta có: ${{\left( \frac{3}{8} \right)}^{5}} > {{\left( \frac{3}{8} \right)}^{3}} > {{\left( \frac{5}{243} \right)}^{3}}$ hay ${{\left( \frac{3}{8} \right)}^{5}} > {{\left( \frac{5}{243} \right)}^{3}}$
h) $A={{10}^{2020}}+\frac{1}{{{10}^{2021}}+1}$
$B={{10}^{2019}}+\frac{1}{{{10}^{2020}}+1}$
$A-B={{10}^{2020}}-{{10}^{2019}}-\left( \frac{1}{{{10}^{2020}}+1}-\frac{1}{{{10}^{2021}}+1} \right)$
$\begin{align}& ={{10}^{2019}}\left( 10-1 \right)-\left( \frac{1}{{{10}^{2020}}+1}-\frac{1}{{{10}^{2021}}+1} \right) \\ & ={{9.10}^{2019}}-\left( \frac{1}{{{10}^{2020}}+1}-\frac{1}{{{10}^{2021}}+1} \right) \\ \end{align}$
$0 < \frac{1}{{{10}^{2020}}+1}-\frac{1}{{{10}^{2021}}+1} < 1\Rightarrow {{9.10}^{2019}}-\left( \frac{1}{{{10}^{2020}}+1}-\frac{1}{{{10}^{2021}}+1} \right) > 0$
hay $A-B > 0$ hay $A > B$
i) ${{5}^{2019}}-{{5}^{2018}}$ và ${{5}^{2020}}-{{5}^{2019}}$
Ta có: ${{5}^{2019}}-{{5}^{2018}}={{5}^{2019}}\left( 1-\frac{1}{5} \right)=\frac{4}{5}{{.5}^{2019}}$
${{5}^{2020}}-{{5}^{2019}}={{5}^{2019}}\left( 5-1 \right)={{4.5}^{2019}}$
$\frac{4}{5}{{.5}^{2019}} < {{4.5}^{2019}}$ hay ${{5}^{2019}}-{{5}^{2018}} < {{5}^{2020}}-{{5}^{2019}}$

So sánh

a) ${{202}^{303}}$ và ${{303}^{202}}$

b) ${{(-32)}^{9}}$và ${{(-18)}^{13}}$

c) ${{5}^{21}}$ và ${{124}^{10}}$

d) ${{(-5)}^{30}}$và ${{(-3)}^{50}}$

e) ${{\left( -\frac{1}{4} \right)}^{8}}$ và ${{\left( \frac{1}{8} \right)}^{50}}$

g) ${{\left( \frac{3}{8} \right)}^{5}}$ và ${{\left( \frac{5}{243} \right)}^{3}}$

h) $A={{10}^{2020}}+{{\left( \frac{1}{10} \right)}^{2021}}+1$ và $B={{10}^{2019}}+{{\left( \frac{1}{10} \right)}^{2020}}+1$

i) ${{5}^{2019}}-{{5}^{2018}}$ và ${{5}^{2020}}-{{5}^{2019}}$

Giải

a) ${{202}^{303}}$ và ${{303}^{202}}$

Ta có: ${{202}^{303}}={{\left( {{202}^{3}} \right)}^{101}}={{\left( {{2}^{3}}{{.101}^{3}} \right)}^{101}}={{\left( {{8.101.101}^{2}} \right)}^{101}}={{\left( {{808.101}^{2}} \right)}^{101}}$

${{303}^{202}}={{\left( {{303}^{2}} \right)}^{101}}={{\left( {{3}^{2}}{{.101}^{2}} \right)}^{101}}={{\left( {{9.101}^{2}} \right)}^{101}}$

Vì $888>9$ nên ${{808.101}^{2}}>{{9.101}^{2}}$ hay ${{\left( {{808.101}^{2}} \right)}^{101}}>{{\left( {{9.101}^{2}} \right)}^{101}}$

Do đó ${{202}^{303}}$ > ${{303}^{202}}$


b) ${{(-32)}^{9}}$và ${{(-18)}^{13}}$

Ta có : ${{(-32)}^{9}}={{(-{{2}^{5}})}^{9}}={{(-2)}^{45}}$

${{(-18)}^{13}}