Kopertina e librit Matematika 12 (me zgjedhje)

Zgjidhja e ushtrimit 1

Zgjidhja e ushtrimit 1 të mësimit 1D në librin Matematika 12 (me zgjedhje) nga shtëpia botuese Mediaprint me autorë Greg Attwood, Jack Barraclough, Ian Bettison, Keith Gallick, Daniel Goldberg, Anne McAteer, Alistair Macpherson, Bronwen Moran, Joe Petran, Keith Bledger, Cong San, Harry Smith, Geoff Staley dhe Dave Wilkins.

Pyetja

Paraqit këto thyesa me anë të thyesave elementare:

  1. 6x−2(x−2)(x+3)\dfrac{6x-2}{(x-2)(x+3)}
  2. 2x+11(x+1)(x+4)\dfrac{2x+11}{(x+1)(x+4)}
  3. −7x−122x(x−4)\dfrac{-7x-12}{2x(x-4)}
  4. 2x−13(2x+1)(x−3)\dfrac{2x-13}{(2x+1)(x-3)}
  5. 6x+6x2−9\dfrac{6x+6}{x^2-9}
  6. 7−3xx2−3x−4\dfrac{7-3x}{x^2-3x-4}
  7. 8−xx2+4x\dfrac{8-x}{x^2+4x}
  8. 2x−14x2+2x−15\dfrac{2x-14}{x^2+2x-15}

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Zgjidhja

  1. 6x−2(x−2)(x+3)≡A(x−2)+B(x+3)≡A(x+3)+B(x−2)(x−2)(x+3)⇒6x−2≡A(x+3)+B(x−2)\dfrac{6x-2}{(x-2)(x+3)} \equiv \dfrac{A}{(x-2)} + \dfrac{B}{(x+3)} \equiv \dfrac{A(x+3)+B(x-2)}{(x-2)(x+3)} \rArr 6x-2 \equiv A(x+3)+B(x-2). Për x=2x = 2 kemi 6×2−2=A(2+3)+B(2−2)⇒10=5A⇒A=26 \times 2 - 2 = A(2+3) + B(2-2) \rArr 10=5A \rArr \boxed{A=2}. Për x=−3x = -3 kemi 6×(−3)−2=A(−3+3)+B(−3−2)⇒−20=B×−5⇒B=46 \times (-3) - 2 = A(-3+3) + B(-3-2) \rArr -20=B \times -5 \rArr \boxed{B=4}. Kështu që 6x−2(x−2)(x+3)≡2(x−2)+4(x+3)\dfrac{6x-2}{(x-2)(x+3)} \equiv \dfrac{2}{(x-2)} + \dfrac{4}{(x+3)}.
  2. 2x+11(x+1)(x+4)≡A(x+1)+B(x+4)≡A(x+4)+B(x+1)(x+1)(x+4)⇒2x+11≡A(x+4)+B(x+1)⇒2x+11≡Ax+4A+Bx+B⇒2x+11≡(A+B)x+(4A+B)\dfrac{2x+11}{(x+1)(x+4)} \equiv \dfrac{A}{(x+1)} + \dfrac{B}{(x+4)} \equiv \dfrac{A(x+4)+B(x+1)}{(x+1)(x+4)} \rArr 2x+11 \equiv A(x+4)+B(x+1) \rArr 2x+11 \equiv Ax+4A+Bx+B \rArr 2x+11 \equiv (A+B)x + (4A+B). Barazojmë koeficientët e xx dhe kemi, 2=A+B2 = A+B. Barazojmë konstantet, 11=4A+B11=4A+B. Me sistem zgjidhim për AA dhe BB dhe kemi A=3\boxed{A=3} dhe B=−1\boxed{B=-1}. Kështu që 2x+11(x+1)(x+4)≡3(x+1)−1(x+4)\dfrac{2x+11}{(x+1)(x+4)} \equiv \dfrac{3}{(x+1)} - \dfrac{1}{(x+4)}.
  3. −7x−122x(x−4)≡A2x+B(x−4)\dfrac{-7x-12}{2x(x-4)} \equiv \dfrac{A}{2x} + \dfrac{B}{(x-4)} ≡\equiv A(x−4)+B×2x2x(x−4)⇒−7x−12≡A(x−4)+2Bx\dfrac{A(x-4)+B \times 2x}{2x(x-4)} \rArr -7x-12 \equiv A(x-4)+2Bx. Për x=4x=4 kemi −7×4−12=A(4−4)+2B×4⇒−40=8B⇒B=−5-7 \times 4 -12 = A(4-4)+2B \times 4 \rArr -40=8B \rArr \boxed{B=-5}. Për x=0x = 0 kemi −7×0−12=A(0−4)+2B×0⇒−12=−4A⇒A=3-7 \times 0 - 12 = A(0-4) + 2B \times 0 \rArr -12 = -4A \rArr \boxed{A=3}. Kështu që −7x−122x(x−4)≡32x−5(x−4)\dfrac{-7x-12}{2x(x-4)} \equiv \dfrac{3}{2x} - \dfrac{5}{(x-4)}.
  4. 2x−13(2x+1)(x−3)≡A(2x+1)+B(x−3)≡A(x−3)+B(2x+1)(2x+1)(x−3)⇒2x−13≡A(x−3)+B(2x+1)\dfrac{2x-13}{(2x+1)(x-3)} \equiv \dfrac{A}{(2x+1)} + \dfrac{B}{(x-3)} \equiv \dfrac{A(x-3)+B(2x+1)}{(2x+1)(x-3)} \rArr 2x-13 \equiv A(x-3) + B(2x+1). Për x=3x = 3 kemi 2×3−13=A(3−3)+B(2×3+1)⇒−7=B×7⇒B=−12 \times 3 - 13 = A(3-3)+B(2\times 3+1) \rArr -7 = B \times 7 \rArr \boxed{B = -1}. Për x=−12x = -\dfrac{1}{2} kemi 2×(−12)−13=A(−12−3)+B(2×(−12)+1)⇒−14=A×−312⇒A=42 \times (-\dfrac{1}{2}) - 13 = A(-\dfrac{1}{2}-3) + B(2 \times (-\dfrac{1}{2})+1) \rArr -14 = A \times -3\dfrac{1}{2} \rArr \boxed{A=4}. Kështu që 2x−13(2x+1)(x−3)≡4(2x+1)−1(x−3)\dfrac{2x-13}{(2x+1)(x-3)} \equiv \dfrac{4}{(2x+1)} - \dfrac{1}{(x-3)}.
  5. Faktorizojmë emëruesin: 6x+6(x+3)(x−3)\dfrac{6x+6}{(x+3)(x-3)}. Tani kemi 6x+6(x+3)(x−3)≡A(x+3)+B(x−3)≡A(x−3)+B(x+3)(x+3)(x−3)⇒6x+6≡A(x−3)+B(x+3)⇒6x+6≡Ax−3A+Bx+3B⇒6x+6≡(A+B)x+(3B−3A)\dfrac{6x+6}{(x+3)(x-3)} \equiv \dfrac{A}{(x+3)} + \dfrac{B}{(x-3)} \equiv \dfrac{A(x-3)+B(x+3)}{(x+3)(x-3)} \rArr 6x+6 \equiv A(x-3)+B(x+3) \rArr 6x+6 \equiv Ax-3A+Bx+3B \rArr 6x+6 \equiv (A+B)x + (3B-3A). Barazojmë koeficientët e xx dhe kemi, 6=A+B6=A+B. Barazojmë konstantet, 6=3B−3A6 = 3B-3A. Me sistem zgjidhim për AA dhe BB dhe kemi A=2\boxed{A=2} dhe B=4\boxed{B=4}. Kështu që, 6x+6x2−9≡2(x+3)+4(x−3)\dfrac{6x+6}{x^2-9} \equiv \dfrac{2}{(x+3)} + \dfrac{4}{(x-3)}.
  6. Faktorizojmë emëruesin: 7−3x(x−4)(x+1)\dfrac{7-3x}{(x-4)(x+1)}. Tani kemi 7−3x(x−4)(x+1)≡A(x−4)+B(x+1)≡A(x+1)+B(x−4)(x−4)(x+1)⇒7−3x≡A(x+1)+B(x−4)\dfrac{7-3x}{(x-4)(x+1)} \equiv \dfrac{A}{(x-4)} + \dfrac{B}{(x+1)} \equiv \dfrac{A(x+1)+B(x-4)}{(x-4)(x+1)} \rArr 7-3x \equiv A(x+1)+B(x-4). Për x=−1x=-1 kemi 7−3×(−1)=A(−1+1)+B(−1−4)⇒10=B×−5⇒B=−27 - 3 \times (-1) = A(-1+1) + B(-1-4) \rArr 10 = B \times -5 \rArr \boxed{B=-2}. Për x=4x = 4 kemi 7−3×4=A(4+1)+B(4−4)⇒−5=A×5⇒A=−17 - 3 \times 4 = A(4+1) + B(4-4) \rArr -5 = A \times 5 \rArr \boxed{A=-1}. Kështu që 7−3xx2−3x−4≡−1(x−4)−2(x+1)\dfrac{7-3x}{x^2-3x-4} \equiv -\dfrac{1}{(x-4)} - \dfrac{2}{(x+1)}.
  7. Faktorizojmë emëruesin: 8−xx(x+4)\dfrac{8-x}{x(x+4)}. Tani kemi 8−xx(x+4)≡Ax+B(x+4)≡A(x+4)+Bxx(x+4)⇒8−x≡A(x+4)+Bx\dfrac{8-x}{x(x+4)} \equiv \dfrac{A}{x} + \dfrac{B}{(x+4)} \equiv \dfrac{A(x+4)+Bx}{x(x+4)} \rArr 8-x \equiv A(x+4) +Bx. Për x=0x=0 kemi 8−0=A(0+4)+B×0⇒8=4A⇒A=28 - 0 = A(0+4)+B \times 0 \rArr 8 = 4A \rArr \boxed{A=2}. Për x=−4x = -4 kemi 8−(−4)=A(−4+4)+B×(−4)⇒12=−4B⇒B=−38-(-4) = A(-4+4)+B \times (-4) \rArr 12 = -4B \rArr \boxed{B=-3}. Kështu që 8−xx2+4x≡2x−3(x+4)\dfrac{8-x}{x^2+4x} \equiv \dfrac{2}{x} - \dfrac{3}{(x+4)}.
  8. Faktorizojmë emëruesin: 2x−14(x+5)(x−3)\dfrac{2x-14}{(x+5)(x-3)}. Tani kemi 2x−14(x+5)(x−3)≡A(x+5)+B(x−3)≡A(x−3)+B(x+5)(x+5)(x−3)⇒2x−14≡A(x−3)+B(x+5)\dfrac{2x-14}{(x+5)(x-3)} \equiv \dfrac{A}{(x+5)} + \dfrac{B}{(x-3)} \equiv \dfrac{A(x-3)+B(x+5)}{(x+5)(x-3)} \rArr 2x-14 \equiv A(x-3)+B(x+5). Për x=3x=3 kemi 2×3−14=A(3−3)+B(3+5)⇒−8=B×8⇒B=−12 \times 3 - 14 = A(3-3) + B(3+5) \rArr -8 = B \times 8 \rArr \boxed{B=-1}. Për x=−5x = -5 kemi 2×(−5)−14=A(−5−3)+B(−5+5)⇒−24=A×(−8)⇒A=32 \times (-5) - 14 = A(-5-3) + B(-5+5) \rArr -24 = A \times (-8) \rArr \boxed{A=3}. Kështu që 2x−14x2+2x−15≡3(x+5)−1(x−3)\dfrac{2x-14}{x^2+2x-15} \equiv \dfrac{3}{(x+5)} - \dfrac{1}{(x-3)}.