Kopertina e librit Matematika 10 - 11: Pjesa II

Zgjidhja e ushtrimit 7

Zgjidhja e ushtrimit 7 të mësimit 1.3A 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

Paraqitni më thjesht shprehjet e mëposhtme.

  1. 35+203\sqrt{5} + \sqrt{20}
  2. 28+57\sqrt{28} + 5\sqrt{7}
  3. 712−2277\sqrt{12} - 2\sqrt{27}
  4. 428−3634\sqrt{28} - 3\sqrt{63}
  5. 50+8−72\sqrt{50} + \sqrt{8} - \sqrt{72}
  6. 356−7126+14003\sqrt{56} - 7\sqrt{126} + \sqrt{1400}

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Zgjidhja

  1. 35+5×4=35+5×4=35+25=553\sqrt{5} + \sqrt{5 \times 4} = 3\sqrt{5} + \sqrt{5} \times \sqrt{4} = 3\sqrt{5} + 2\sqrt{5} = 5\sqrt{5}
  2. 7×4+57=7×4+57=27+57=77\sqrt{7 \times 4} + 5\sqrt{7} = \sqrt{7} \times \sqrt{4} + 5\sqrt{7} = 2\sqrt{7} + 5\sqrt{7} = 7\sqrt{7}
  3. 74×3−29×3=7×4×3−2×9×3=143−63=837\sqrt{4 \times 3} - 2\sqrt{9 \times 3} = 7 \times \sqrt{4} \times \sqrt{3} - 2 \times \sqrt{9} \times \sqrt{3} = 14\sqrt{3} - 6\sqrt{3} = 8\sqrt{3}
  4. 47×4−39×7=4×7×4−3×9×7=87−97=−74\sqrt{7 \times 4} - 3\sqrt{9 \times 7} = 4 \times \sqrt{7} \times \sqrt{4} - 3 \times \sqrt{9} \times \sqrt{7} = 8\sqrt{7} - 9\sqrt{7} = -\sqrt{7}
  5. 25×2+4×2−36×2=25×2+4×2−36×2=52+22−62=2\sqrt{25 \times 2} + \sqrt{4 \times 2} - \sqrt{36 \times 2} = \sqrt{25} \times \sqrt{2} + \sqrt{4} \times \sqrt{2} - \sqrt{36} \times \sqrt{2} = 5\sqrt{2} + 2\sqrt{2} - 6\sqrt{2} = \sqrt{2}
  6. 314×4−714×9+100×14=3×14×4−7×14×9+100×14=614−2114+1014=−5143\sqrt{14 \times 4} - 7\sqrt{14 \times 9} + \sqrt{100 \times 14} = 3 \times \sqrt{14} \times \sqrt{4} - 7 \times \sqrt{14} \times \sqrt{9} + \sqrt{100} \times \sqrt{14} = 6\sqrt{14} - 21\sqrt{14} + 10\sqrt{14} = -5\sqrt{14}