Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 13

Zgjidhja e ushtrimit 13 të mësimit 6.4A në librin Matematika 10 - 11: Pjesa I nga shtëpia botuese Botime Pegi me autorë Steve Fearnley, June Haighton, Steve Lomax, Peter Mullarkey, James Nicholson dhe Matt Nixon.

Pyetja

Faktorizoni shprehjet algjebrike të mëposhtme.

  1. x2+4xy+4y2x^2+4xy+4y^2
  2. 16x2−15xy+9y216x^2-15xy+9y^2
  3. 5xy+10x2y25xy+10x^2y^2
  4. 10−3x−x210-3x-x^2
  5. 10x+3−8x210x+3-8x^2
  6. 15xy+2y−3x2y15xy+2y-3x^2y
  7. x4−7x2+12x^4-7x^2+12
  8. 2x4y4+x2y2−62x^4y^4+x^2y^2-6

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Zgjidhja

  1. x2+2xy+2xy+4y2=x(x+2y)+2y(x+2y)=(x+2y)(x+2y)x^2 + 2xy + 2xy + 4y^2 = x(x+2y)+2y(x+2y)=(x+2y)(x+2y) == (x+2y)2(x+2y)^2
  2. Nuk mund të faktorizohet.
  3. 5xy(1+2xy)5xy(1+2xy)
  4. −x2−3x+10=−x2+2x−5x+10=−x(x−2)−5(x−2)=(−x−5)(x−2)=−(x+5)(x−2)-x^2-3x+10=-x^2+2x-5x+10=-x(x-2)-5(x-2)=(-x-5)(x-2)=-(x+5)(x-2)
  5. −8x2+10x+3=−8x2−2x+12x+3=−2x(4x+1)+3(4x+1)=(−2x+3)(4x+1)=−(2x−3)(4x+1)-8x^2+10x+3=-8x^2-2x+12x+3=-2x(4x+1)+3(4x+1)=(-2x+3)(4x+1)=-(2x-3)(4x+1)
  6. −y(3x2−15x−2)-y(3x^2-15x-2)
  7. x4−4x2−3x2+12=x2(x2−4)−3(x2−4)=(x2−3)(x2−4)x^4-4x^2-3x^2+12=x^2(x^2-4)-3(x^2-4)=(x^2-3)(x^2-4) == (x2−3)(x2−22)=(x2−3)(x+2)(x−2)(x^2-3)(x^2-2^2)=(x^2-3)(x+2)(x-2)
  8. 2x4y4+4x2y2−3x2y2−6=2x2y2(x2y2+2)−3(x2y2+2)=(2x2y2−3)(x2y2+2)2x^4y^4+4x^2y^2-3x^2y^2-6=2x^2y^2(x^2y^2+2)-3(x^2y^2+2)=(2x^2y^2-3)(x^2y^2+2)