Ontbinden in factoren (1)

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Ontbind in factoren door gebruik te maken van merkwaardige producten

  1. \(q^{10}-169s^2\)
  2. \(p^2-10p+25\)
  3. \(196b^{10}+308b^5p+121p^2\)
  4. \(100q^{10}-180q^5+81\)
  5. \(-81b^2+196\)
  6. \(81p^2-4\)
  7. \(-225q^2+121\)
  8. \(25b^2+90b+81\)
  9. \(196y^{6}+140y^3+25\)
  10. \(1-169p^{16}\)
  11. \(49-100y^{10}\)
  12. \(25a^2-140a+196\)

Ontbind in factoren door gebruik te maken van merkwaardige producten

Verbetersleutel

  1. \(q^{10}-169s^2=(q^5+13s)(q^5-13s)\)
  2. \(p^2-10p+25=(p-5)^2\)
  3. \(196b^{10}+308b^5p+121p^2=(14b^5+11p)^2\)
  4. \(100q^{10}-180q^5+81=(10q^5-9)^2\)
  5. \(-81b^2+196=(14-9b)(14+9b)\)
  6. \(81p^2-4=(9p+2)(9p-2)\)
  7. \(-225q^2+121=(11-15q)(11+15q)\)
  8. \(25b^2+90b+81=(5b+9)^2\)
  9. \(196y^{6}+140y^3+25=(14y^3+5)^2\)
  10. \(1-169p^{16}=(1-13p^8)(1+13p^8)\)
  11. \(49-100y^{10}=(7-10y^5)(7+10y^5)\)
  12. \(25a^2-140a+196=(5a-14)^2\)
Oefeningengenerator wiskundeoefeningen.be 2026-01-11 16:14:30
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