Més exercicis sobre radicals

13 11 2008

Alumnat d’EPPA,

Aquí teniu les solucions als problemes 5 i 6 del model d’examen de la unitat 1, exercicis que no hem pogut resoldre a classe.

Problema 5.- Calculeu i simplifiqueu:

  • 3\sqrt{72}-\sqrt{18}+5\sqrt{2}+\sqrt{50}-2\sqrt{8} =
    = 3\sqrt{2^3\cdot 3^2} - \sqrt{2\cdot 3^2}+5\sqrt{2}+\sqrt{2\cdot 5^2} -2\sqrt{2^3}=
    = 3\cdot 2 \cdot 3 \sqrt{2} - 3\sqrt{2} + 5\sqrt{2} +5\sqrt{2} -2\cdot 2\sqrt{2}=
    = 18\sqrt{2}-3\sqrt{2}+5\sqrt{2}+5\sqrt{2} -4\sqrt{2}= 21\sqrt{2}
  • 9\sqrt{27}+2\sqrt{3}-8\sqrt{300}-4\sqrt{3} =
    = 9\sqrt{3^3}+2\sqrt{3}-8\sqrt{3\cdot 10^2}-4\sqrt{3}=
    = 9\cdot 3 \sqrt{3} + 2\sqrt{3} - 8 \cdot 10 \sqrt{3}-4\sqrt{3} =
    = 27\sqrt{3} + 2\sqrt{3} -80 \sqrt{3} -4\sqrt{3} = -55\sqrt{3}
  • \left(3+\sqrt{2} \right)\left(3-\sqrt{2} \right) = 3^2-\left(\sqrt{2}\right)^2=9-2=7
  • \left(\sqrt{3}+ 2 \right) \left(\sqrt{5}+\sqrt{3} \right) =
    = \sqrt{3}\cdot \sqrt{5} + \left(\sqrt{3}\right)^2 +2\sqrt{5} + 2\sqrt{3}=
    = \sqrt{15} +3 +2\sqrt{5} + 2\sqrt{3}=2\sqrt{3}+2\sqrt{5}+\sqrt{15}+3

Problema 6.- Racionalitzeu i simplifiqueu:

  • \dfrac{4}{\sqrt{3}}=\dfrac{4}{\sqrt{3}} \cdot \dfrac{\sqrt{3}}{\sqrt{3}}=\dfrac{4\sqrt{3}}{\left(\sqrt{3}\right)^2}=\dfrac{4\sqrt{3}}{3}
  • \dfrac{\sqrt{2}}{\sqrt{5}-\sqrt{3}} = \dfrac{\sqrt{2}}{\sqrt{5}-\sqrt{3}}\cdot \dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}+\sqrt{3}}=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{\left(\sqrt{5}\right)^2-\left(\sqrt{3}\right)^2}=
    =\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{5-3}=\dfrac{\sqrt{2}\left(\sqrt{5}+\sqrt{3}\right)}{2}

Si teniu qualsevol dubte, no dubteu a escriure un comentari.

Estudieu!!

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