Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 11

Zgjidhja e ushtrimit 11 të mësimit Vlerësim 2 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

Shkruani secilën nga shprehjet e mëposhtme si një thyesë të vetme në trajtë të thjeshtuar.

  1. z−36+2z+18\dfrac{z-3}{6} + \dfrac{2z+1}{8}
  2. 7−2y5−6y−27\dfrac{7-2y}{5} - \dfrac{6y-2}{7}
  3. 4x+1+3x+2\dfrac{4}{x+1} + \dfrac{3}{x+2}
  4. 72p+3−43p−2\dfrac{7}{2p+3} - \dfrac{4}{3p-2}
  5. 2aa−2+3a2a+9\dfrac{2a}{a-2} + \dfrac{3a}{2a+9}
  6. 3(2x+1)2−4(2x+1)3\dfrac{3}{(2x+1)^2} - \dfrac{4}{(2x+1)^3}

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Zgjidhja

  1. 4(z−3)+3(2z+1)24=4z−12+6z+324=10z−924\dfrac{4(z-3)+3(2z+1)}{24} = \dfrac{4z-12+6z+3}{24} = \dfrac{10z-9}{24}
  2. 7(7−2y)−5(6y−2)35=49−14y−30y+1035=59−44y35\dfrac{7(7-2y)-5(6y-2)}{35} = \dfrac{49-14y-30y+10}{35} = \dfrac{59-44y}{35}
  3. 4(x+2)+3(x+1)(x+1)(x+2)=4x+8+3x+3(x+1)(x+2)=7x+11(x+1)(x+2)\dfrac{4(x+2)+3(x+1)}{(x+1)(x+2)} = \dfrac{4x+8+3x+3}{(x+1)(x+2)} = \dfrac{7x+11}{(x+1)(x+2)}
  4. 7(3p−2)−4(2p+3)(2p+3)(3p−2)=21p−14−8p−12(2p+3)(3p−2)=13p−26(2p+3)(3p−2)=13(p−2)(2p+3)(3p−2)\dfrac{7(3p-2)-4(2p+3)}{(2p+3)(3p-2)} = \dfrac{21p-14-8p-12}{(2p+3)(3p-2)} = \dfrac{13p-26}{(2p+3)(3p-2)} = \dfrac{13(p-2)}{(2p+3)(3p-2)}
  5. 2a(2a+9)+3a(a−2)(a−2)(2a+9)=4a2+18a+3a2−6a(a−2)(2a+9)=7a2+12a(a−2)(2a+9)=a(7a+12)(a−2)(2a+9)\dfrac{2a(2a+9)+3a(a-2)}{(a-2)(2a+9)} = \dfrac{4a^2+18a+3a^2-6a}{(a-2)(2a+9)} = \dfrac{7a^2+12a}{(a-2)(2a+9)} = \dfrac{a(7a+12)}{(a-2)(2a+9)}
  6. 3(2x+1)−4(2x+1)3=6x+3−4(2x+1)3=6x−1(2x+1)3\dfrac{3(2x+1)-4}{(2x+1)^3} = \dfrac{6x+3-4}{(2x+1)^3} = \dfrac{6x-1}{(2x+1)^3}