Kopertina e librit Matematika 10 - 11: Pjesa II

Zgjidhja e ushtrimit 8

Zgjidhja e ushtrimit 8 të mësimit Përsëritje 1 në librin Matematika 10 - 11: Pjesa II nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Eleminoni rrënjët nga emëruesi i thyesave të mëposhtme.

  1. 17\dfrac{1}{\sqrt{7}}
  2. 36\dfrac{3}{\sqrt{6}}
  3. 525\dfrac{5}{2\sqrt{5}}
  4. 4+66\dfrac{4+\sqrt{6}}{\sqrt{6}}
  5. 32−5128−42\dfrac{3\sqrt{2}-5}{\sqrt{128}-4\sqrt{2}}
  6. 43+5263+27\dfrac{4\sqrt{3}+5\sqrt{2}}{6\sqrt{3}+\sqrt{27}}

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Zgjidhja

  1. 17×77\dfrac{1}{\sqrt{7}} \times \dfrac{\sqrt{7}}{\sqrt{7}} == 77\dfrac{\sqrt{7}}{7}
  2. 36×66=366=62\dfrac{3}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} = \dfrac{3\sqrt{6}}{6} = \dfrac{\sqrt{6}}{2}
  3. 525×55=552×5=52\dfrac{5}{2\sqrt{5}} \times \dfrac{\sqrt{5}}{\sqrt{5}} = \dfrac{\cancel5\sqrt{5}}{2 \times \cancel5} = \dfrac{\sqrt{5}}{2}
  4. 4+66×66=46+66=2(26+3)6=26+33\dfrac{4+\sqrt{6}}{\sqrt{6}} \times \dfrac{\sqrt{6}}{\sqrt{6}} = \dfrac{4\sqrt{6}+6}{6} = \dfrac{2(2\sqrt{6}+3)}{6} = \dfrac{2\sqrt{6}+3}{3}
  5. 32−564×2−42=32−582−42=32−542⇒32−542×22=3×2−524×2=6−528\dfrac{3\sqrt{2}-5}{\sqrt{64 \times 2}-4\sqrt{2}} = \dfrac{3\sqrt{2}-5}{8\sqrt{2}-4\sqrt{2}} = \dfrac{3\sqrt{2}-5}{4\sqrt{2}} \rArr \dfrac{3\sqrt{2}-5}{4\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{3\times 2 - 5\sqrt{2}}{4 \times 2} = \dfrac{6-5\sqrt{2}}{8}
  6. 43+5263+33⇒43+5293×33=4×3+52×39×3=12+5627\dfrac{4\sqrt{3}+5\sqrt{2}}{6\sqrt{3}+3\sqrt{3}} \rArr \dfrac{4\sqrt{3}+5\sqrt{2}}{9\sqrt{3}} \times \dfrac{\sqrt{3}}{\sqrt{3}} = \dfrac{4\times 3+5\sqrt{2 \times 3}}{9 \times 3} = \dfrac{12+5\sqrt{6}}{27}