Category: Ainda os números

• Enunciado
• Resolução

Determina:

1. ${{3}^{3}}\div {{3}^{6}}+{{\left( {{3}^{2}} \right)}^{-1}}$
    
2. ${{10}^{0}}+{{7}^{-1}}\times {{7}^{2}}\div {{7}^{-3}}$
    
3. ${{({{4}^{0}}-{{4}^{-1}}+{{4}^{-2}})}^{-6}}\div {{\left( \frac{16}{13} \right)}^{5}}$

Resolução >> Resolução

1.  
   \[{{3}^{3}}\div {{3}^{6}}+{{\left( {{3}^{2}} \right)}^{-1}}={{3}^{-3}}+{{3}^{-2}}={{\left( \frac{1}{3} \right)}^{3}}+{{\left( \frac{1}{3} \right)}^{2}}=\frac{1}{27}+\frac{1}{9}=\frac{1}{27}+\frac{3}{27}=\frac{4}{27}\]
2.  
   \[{{10}^{0}}+{{7}^{-1}}\times {{7}^{2}}\div {{7}^{-3}}=1+{{7}^{1}}\div {{7}^{-3}}=1+{{7}^{4}}=1+2401=2402\]
3.  
   \[\begin{array}{*{35}{l}}
      {{({{4}^{0}}-{{4}^{-1}}+{{4}^{-2}})}^{-6}}\div {{\left( \frac{16}{13} \right)}^{5}} & = & {{(1-\frac{1}{4}+\frac{1}{16})}^{-6}}\div {{\left( \frac{16}{13} \right)}^{5}}  \\
      {} & = & {{(\frac{16}{16}-\frac{4}{16}+\frac{1}{16})}^{-6}}\div {{\left( \frac{16}{13}

• Enunciado
• Resolução

Calcula o valor numérico das expressões, utilizando, sempre que possível, as regras das potências:

1. ${{(0,4)}^{-2}}\times {{4}^{-2}}$
    
2. ${{(0,3)}^{-5}}\div {{(-0,3)}^{-2}}$
    
3. ${{\left( -\frac{1}{4} \right)}^{-3}}\times {{\left( -\frac{1}{4} \right)}^{2}}$
    
4. ${{(-100)}^{-2}}\times {{(0,03)}^{-2}}$
    
5. ${{(0,125)}^{-4}}\div {{\left( \frac{1}{2} \right)}^{-4}}$

Resolução >> Resolução

1.  
   \[{{(0,4)}^{-2}}\times {{4}^{-2}}={{1,6}^{-2}}={{\left( \frac{16}{10} \right)}^{-2}}={{\left( \frac{8}{5} \right)}^{-2}}={{\left( \frac{5}{8} \right)}^{2}}=\frac{25}{64}\]
2.  
   \[{{(0,3)}^{-5}}\div {{(-0,3)}^{-2}}={{(0,3)}^{-5}}\div {{(+0,3)}^{-2}}={{(0,3)}^{-5-(-2)}}={{\left( \frac{3}{10} \right)}^{-3}}={{\left( \frac{10}{3}

• Enunciado
• Resolução

Escreve como potência de expoente 2 a soma: $1+{{3}^{-1}}+{{3}^{-2}}+{{3}^{-3}}+{{3}^{-4}}$

Resolução >> Resolução

\[\begin{array}{*{35}{l}}
   1+{{3}^{-1}}+{{3}^{-2}}+{{3}^{-3}}+{{3}^{-4}} & = & 1+\frac{1}{3}+\frac{1}{9}+\frac{1}{27}+\frac{1}{81}  \\
   {} & = & \frac{81}{81}+\frac{27}{81}+\frac{9}{81}+\frac{3}{81}+\frac{1}{81}  \\
   {} & = & \frac{121}{81}  \\
   {} & = & {{\left( \frac{11}{9} \right)}^{2}}  \\
\end{array}\]

• Enunciado
• Resolução

1. ${{6}^{0}}$
    
2. ${{(-5)}^{-1}}$
    
3. ${{\left( \frac{1}{8} \right)}^{-1}}$
    
4. ${{\left( \frac{7}{4} \right)}^{-1}}$
    
5. ${{\left( -\frac{12}{5} \right)}^{-2}}$
    
6. ${{\left( -\frac{6}{7} \right)}^{-3}}$
    
7. ${{\left( \frac{3}{10} \right)}^{-4}}$
    
8. ${{0,1}^{-1}}$

Resolução >> Resolução

1.  
   \[{{6}^{0}}=1\]
2.  
   \[{{(-5)}^{-1}}={{\left( -\frac{1}{5} \right)}^{1}}=-\frac{1}{5}\]
3.  
   \[{{\left( \frac{1}{8} \right)}^{-1}}={{8}^{1}}=8\]
4.  
   \[{{\left( \frac{7}{4} \right)}^{-1}}={{\left( \frac{4}{7} \right)}^{1}}=\frac{4}{7}\]
5.  
   \[{{\left( -\frac{12}{5} \right)}^{-2}}={{\left( -\frac{5}{12} \right)}^{2}}=\frac{25}{144}\]
6.  
   \[{{\left( -\frac{6}{7} \right)}^{-3}}={{\left( -\frac{7}{6} \right)}^{3}}=-\frac{343}{216}\]
7.  
   \[{{\left(