Calcula o valor das expressões
Números reais: Matematicamente Falando 8 - Pág. 33 Ex. 1
Calcula o valor das expressões, tendo em atenção as propriedades das operações.
- \(2\sqrt 3 + 4\sqrt 3 – 5\sqrt 3 \)
- \({\left( {\sqrt 2 + 2} \right)^2}\)
- \(\frac{1}{3}\pi – \pi + 3\pi \)
- \(\left( {5 – \sqrt 5 } \right)\left( {5 + \sqrt 5 } \right)\)
- \({\left( {\sqrt 7 – 1} \right)^2}\)
- \(\left( {2 + \sqrt 3 } \right)\left( {7 – \sqrt 3 } \right)\)
- \(\sqrt 5 – \sqrt 6 + 2\sqrt 5 – 2\sqrt 6 \)
- \(\sqrt {\frac{9}{{25}}} – \frac{4}{5} + \sqrt 2 \)
- \(\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 – 1} \right)\)
- \(\left( {\sqrt {10} – \sqrt {11} } \right)\left( {\sqrt {10} – \sqrt {11} } \right)\)
- \({\left( {2\sqrt 3 } \right)^2}\)
- \({\left( {3\sqrt 2 } \right)^2}\)
- \({\left( {3\sqrt 5 } \right)^2}\)
Tendo em atenção as propriedades das operações, apresenta-se o cálculo das expressões dadas:
| Alínea | Cálculo da expressão |
| a) | \(2\sqrt 3 + 4\sqrt 3 – 5\sqrt 3 = \left( {2 + 4 – 5} \right) \times \sqrt 3 = \sqrt 3 \) |
| b) | \({\left( {\sqrt 2 + 2} \right)^2} = \left( {\sqrt 2 + 2} \right)\left( {\sqrt 2 + 2} \right) = {\left( {\sqrt 2 } \right)^2} + 2\sqrt 2 + 2\sqrt 2 + 4 = 2 + 4\sqrt 2 + 4 = 6 + 4\sqrt 2 \) |
| c) | \(\frac{1}{3}\pi – \pi + 3\pi = \left( {\frac{1}{3} – 1 + 3} \right)\pi = \frac{7}{3}\pi \) |
| d) | \(\left( {5 – \sqrt 5 } \right)\left( {5 + \sqrt 5 } \right) = 25 + 5\sqrt 5 – 5\sqrt 5 – {\left( {\sqrt 5 } \right)^2} = 25 – 5 = 20\) |
| e) | \({\left( {\sqrt 7 – 1} \right)^2} = \left( {\sqrt 7 – 1} \right)\left( {\sqrt 7 – 1} \right) = {\left( {\sqrt 7 } \right)^2} – \sqrt 7 – \sqrt 7 + 1 = 7 – 2\sqrt 7 + 1 = 8 – 2\sqrt 7 \) |
| f) | \(\left( {2 + \sqrt 3 } \right)\left( {7 – \sqrt 3 } \right) = 14 – 2\sqrt 3 + 7\sqrt 3 – {\left( {\sqrt 3 } \right)^2} = 14 + 5\sqrt 3 – 3 = 11 + 5\sqrt 3 \) |
| g) | \(\sqrt 5 – \sqrt 6 + 2\sqrt 5 – 2\sqrt 6 = \left( {\sqrt 5 + 2\sqrt 5 } \right) + \left( { – \sqrt 6 – 2\sqrt 6 } \right) = 3\sqrt 5 – 3\sqrt 6 \) |
| h) | \(\sqrt {\frac{9}{{25}}} – \frac{4}{5} + \sqrt 2 = \frac{3}{5} – \frac{4}{5} + \sqrt 2 = – \frac{1}{5} + \sqrt 2 \) |
| i) | \(\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 – 1} \right) = {\left( {\sqrt 2 } \right)^2} – \sqrt 2 + \sqrt 2 – 1 = 2 – 1 = 1\) |
| j) | \(\left( {\sqrt {10} – \sqrt {11} } \right)\left( {\sqrt {10} + \sqrt {11} } \right) = {\left( {\sqrt {10} } \right)^2} + \sqrt {10} \times \sqrt {11} – \sqrt {10} \times \sqrt {11} – {\left( {\sqrt {11} } \right)^2} = 10 – 11 = – 1\) |
| k) | \({\left( {2\sqrt 3 } \right)^2} = {2^2} \times {\left( {\sqrt 3 } \right)^2} = 4 \times 3 = 12\) |
| l) | \({\left( {3\sqrt 2 } \right)^2} = 3\sqrt 2 \times 3\sqrt 2 = \left( {3 \times 3} \right) \times \left( {\sqrt 2 \times \sqrt 2 } \right) = 9 \times 2 = 18\) |
| m) | \({\left( {3\sqrt 5 } \right)^2} = {3^2} \times {\left( {\sqrt 5 } \right)^2} = 9 \times 5 = 45\) |














