Kopertina e librit Matematika 12

Zgjidhja e ushtrimit 2

Zgjidhja e ushtrimit 2 të mësimit 7A në librin Matematika 12 nga shtëpia botuese Mediaprint me autorë Greg Attwood, Jack Barraclough, Ian Bettison, Alistair Macpherson, Bronwen Moran, Su Nicholson, Diane Oliver, Joe Petran, Keith Bledger, Harry Smith, Geoff Staley, Robert Ward-Penny dhe Dave Wilkins.

Pyetja

Thjeshto këto thyesa plotësisht:

  1. (x+3)(x−2)(x−2)\dfrac{(x + 3)(x - 2)}{(x - 2)}
  2. (x+4)(3x−1)(3x−1)\dfrac{(x + 4)(3x - 1)}{(3x - 1)}
  3. (x+3)2(x+3)\dfrac{(x + 3)^2}{(x + 3)}
  4. x2+10x+21(x+3)\dfrac{x^2 + 10x + 21}{(x + 3)}
  5. x2+9x+20(x+4)\dfrac{x^2 + 9x + 20}{(x + 4)}
  6. x2+x−12(x−3)\dfrac{x^2 + x - 12}{(x - 3)}
  7. x2+x−20x2+2x−15\dfrac{x^2 + x - 20}{x^2 + 2x - 15}
  8. x2+3x+2x2+5x+4\dfrac{x^2 + 3x + 2}{x^2 + 5x + 4}
  9. x2+x−12x2−9x+18\dfrac{x^2 + x - 12}{x^2 - 9x + 18}
  10. 2x2+7x+6(x−5)(x+2)\dfrac{2x^2 + 7x + 6}{(x - 5)(x + 2)}
  11. 2x2+9x−18(x+6)(x+1)\dfrac{2x^2 + 9x - 18}{(x + 6)(x + 1)}
  12. 3x2−7x+2(3x−1)(x+2)\dfrac{3x^2 - 7x + 2}{(3x - 1)(x + 2)}
  13. 2x2+3x+1x2−x−2\dfrac{2x^2 + 3x + 1}{x^2 - x - 2}
  14. x2+6x+83x2+7x+2\dfrac{x^2 + 6x + 8}{3x^2 + 7x + 2}
  15. 2x2−5x−32x2−9x+9\dfrac{2x^2 - 5x - 3}{2x^2 - 9x + 9}

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Zgjidhja

  1. (x+3)(x−2)(x−2)=x+3\dfrac{(x + 3)\cancel{(x - 2)}}{\cancel{(x - 2)}} = x + 3
  2. (x+4)(3x−1)(3x−1)=x+4\dfrac{(x + 4)\cancel{(3x - 1)}}{\cancel{(3x - 1)}} = x + 4
  3. (x+3)(x+3)(x+3)=x+3\dfrac{(x + 3)\cancel{\cancel{(x + 3)}}}{\cancel{(x + 3)}} = x + 3
  4. x2+3x+7x+21(x+3)=x(x+3)+7(x+3)(x+3)=(x+7)(x+3)(x+3)=x+7\dfrac{x^2 + 3x + 7x + 21}{(x + 3)} = \dfrac{x(x + 3) + 7(x + 3)}{(x + 3)} = \dfrac{(x + 7)\cancel{(x + 3)}}{\cancel{(x + 3)}} = x + 7
  5. x2+4x+5x+20(x+4)=x(x+4)+5(x+4)(x+4)=(x+5)(x+4)(x+4)=x+5\dfrac{x^2 + 4x + 5x + 20}{(x + 4)} = \dfrac{x(x + 4) + 5(x + 4)}{(x + 4)} = \dfrac{(x + 5)\cancel{(x + 4)}}{\cancel{(x + 4)}} = x + 5
  6. x2+4x−3x−12(x−3)=x(x+4)−3(x+4)(x−3)=(x−3)(x+4)(x−3)=x+4\dfrac{x^2 + 4x - 3x - 12}{(x - 3)} = \dfrac{x(x + 4) - 3(x + 4)}{(x - 3)} = \dfrac{\cancel{(x - 3)}(x + 4)}{\cancel{(x - 3)}} = x + 4
  7. x2+5x−4x−20x2+5x−3x−15=x(x+5)−4(x+5)x(x+5)−3(x+5)=(x+5)(x−4)(x+5)(x−3)=x−4x−3\dfrac{x^2 + 5x - 4x - 20}{x^2 + 5x - 3x - 15} = \dfrac{x(x + 5) - 4(x + 5)}{x(x + 5) - 3(x + 5)} = \dfrac{\cancel{(x + 5)}(x - 4)}{\cancel{(x + 5)}(x - 3)} = \dfrac{x - 4}{x - 3}
  8. x2+x+2x+2x2+x+4x+4=x(x+1)+2(x+1)x(x+1)+4(x+1)=(x+2)(x+1)(x+4)(x+1)=x+2x+4\dfrac{x^2 + x + 2x + 2}{x^2 + x + 4x + 4} = \dfrac{x(x + 1) + 2(x + 1)}{x(x + 1) + 4(x + 1)} = \dfrac{(x + 2)\cancel{(x + 1)}}{(x + 4)\cancel{(x + 1)}} = \dfrac{x + 2}{x + 4}
  9. x2+4x−3x−12x2−6x−3x+18=x(x+4)−3(x+4)x(x−6)−3(x−6)=(x+4)(x−3)(x−6)(x−3)=x+4x−6\dfrac{x^2 + 4x - 3x - 12}{x^2 - 6x - 3x + 18} = \dfrac{x(x + 4) - 3(x + 4)}{x(x - 6) - 3(x - 6)} = \dfrac{(x + 4)\cancel{(x - 3)}}{(x - 6)\cancel{(x - 3)}} = \dfrac{x + 4}{x - 6}
  10. 2x2+4x+3x+6(x−5)(x+2)=2x(x+2)+3(x+2)(x−5)(x+2)=(2x+3)(x+2)(x+5)(x−2)=2x+3x−5\dfrac{2x^2 + 4x + 3x + 6}{(x - 5)(x + 2)} = \dfrac{2x(x + 2) + 3(x + 2)}{(x - 5)(x + 2)} = \dfrac{(2x + 3)\cancel{(x + 2)}}{(x + 5)\cancel{(x - 2)}} = \dfrac{2x + 3}{x - 5}
  11. 2x2+12x−3x−18(x+6)(x+1)=2x(x+6)−3(x+6)(x+6)(x+1)=(2x−3)(x+6)(x+6)(x+1)=2x−3x+1\dfrac{2x^2 + 12x - 3x - 18}{(x + 6)(x + 1)} = \dfrac{2x(x + 6) - 3(x + 6)}{(x + 6)(x + 1)} = \dfrac{(2x - 3)\cancel{(x + 6)}}{\cancel{(x + 6)}(x + 1)} = \dfrac{2x - 3}{x + 1}
  12. 3x2−6x−x+2(3x−1)(x+2)=3x(x−2)−(x−2)(3x−1)(x+2)=(3x−1)(x−2)(3x−1)(x+2)=x−2x+2\dfrac{3x^2 - 6x - x + 2}{(3x - 1)(x + 2)} = \dfrac{3x(x - 2) - (x - 2)}{(3x - 1)(x + 2)} = \dfrac{\cancel{(3x - 1)}(x - 2)}{\cancel{(3x - 1)}(x + 2)} = \dfrac{x - 2}{x + 2}
  13. 2x2+2x+x+1x2−2x+x−2=2x(x+1)+(x+1)x(x−2)+(x−2)=(2x+1)(x+1)(x−2)(x+1)=2x+1x−2\dfrac{2x^2 + 2x + x + 1}{x^2 - 2x + x - 2} = \dfrac{2x(x + 1) + (x + 1)}{x(x - 2) + (x - 2)} = \dfrac{(2x + 1)\cancel{(x + 1)}}{(x - 2)\cancel{(x + 1)}} = \dfrac{2x + 1}{x - 2}
  14. x2+2x+4x+83x2+6x+x+2=x(x+2)+4(x+2)3x(x+2)+(x+2)=(x+4)(x+2)(3x+1)(x+2)=x+43x+1\dfrac{x^2 + 2x + 4x + 8}{3x^2 + 6x + x + 2} = \dfrac{x(x + 2) + 4(x + 2)}{3x(x + 2) + (x + 2)} = \dfrac{(x + 4)\cancel{(x + 2)}}{(3x + 1)\cancel{(x + 2)}} = \dfrac{x + 4}{3x + 1}
  15. 2x2−6x+x−32x2−6x−3x+9=2x(x−3)+(x−3)2x(x−3)−3(x−3)=(2x+1)(x−3)(2x−3)(x−3)=2x+12x−3\dfrac{2x^2 - 6x + x - 3}{2x^2 - 6x - 3x + 9} = \dfrac{2x(x - 3) + (x - 3)}{2x(x - 3) - 3(x - 3)} = \dfrac{(2x + 1)\cancel{(x - 3)}}{(2x - 3)\cancel{(x - 3)}} = \dfrac{2x + 1}{2x - 3}