Kopertina e librit Matematika 10-11 (fletore pune)

Zgjidhja e ushtrimit 6

Zgjidhja e ushtrimit 6 të mësimit 2.4 - Thyesat algjebrike në librin Matematika 10-11 (fletore pune) nga shtëpia botuese Botime Pegi me autorë Clare Pass.

Pyetja

Thjeshtoni thyesat e mëposhtme.

  1. x2+2x3x−3×6x−62x+4\dfrac{x^2+2x}{3x-3} \times \dfrac{6x-6}{2x+4}
  2. 2x2−6x2x2+5x:3x−912x+30\dfrac{2x^2-6x}{2x^2+5x} : \dfrac{3x-9}{12x+30}
  3. y2−3y3y+6×4y+8(y−3)2\dfrac{y^2-3y}{3y+6} \times \dfrac{4y+8}{(y-3)^2}

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Zgjidhja

  1. x(x+2)3(x−1)×6(x−1)2(x+2)=x(x+2)3(x−1)×6(x−1)2(x+2)=6x6=x\dfrac{x(x+2)}{3(x-1)} \times \dfrac{6(x-1)}{2(x+2)} = \dfrac{x\cancel{(x+2)}}{3\cancel{(x-1)}} \times \dfrac{6\cancel{(x-1)}}{2\cancel{(x+2)}} = \dfrac{6x}{6} = x
  2. 2x(x−3)x(2x+5)×6(2x+5)3(x−3)=2x(x−3)x(2x+5)×6(2x+5)3(x−3)=12x3x=4\dfrac{2x(x-3)}{x(2x+5)} \times \dfrac{6(2x+5)}{3(x-3)} = \dfrac{2x\cancel{(x-3)}}{x\cancel{(2x+5)}} \times \dfrac{6\cancel{(2x+5)}}{3\cancel{(x-3)}} = \dfrac{12x}{3x} = 4
  3. y(y−3)3(y+2)×4(y+2)(y−3)2=y(y−3)3(y+2)×4(y+2)(y−3)2=4y3(y−3)=4y3y−9\dfrac{y(y-3)}{3(y+2)} \times \dfrac{4(y+2)}{(y-3)^2} = \dfrac{y\cancel{(y-3)}}{3\cancel{(y+2)}} \times \dfrac{4\cancel{(y+2)}}{(y-3)\cancel{^2}} = \dfrac{4y}{3(y-3)} = \dfrac{4y}{3y-9}