Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 13

Zgjidhja e ushtrimit 13 të mësimit 2.4A 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

Paraqitni më thjesht shprehjet e mëposhtme.

  1. 5x+2×2x\dfrac{5}{x+2} \times \dfrac{2}{x}
  2. 3x×xx−1\dfrac{3}{x} \times \dfrac{x}{x-1}
  3. 3x+2×3x+62x+3\dfrac{3}{x+2} \times \dfrac{3x+6}{2x+3}
  4. xx+2×x2+2xx2\dfrac{x}{x+2} \times \dfrac{x^2+2x}{x^2}
  5. xx+2:x2\dfrac{x}{x+2} : \dfrac{x}{2}
  6. 5y−1015:y−23y\dfrac{5y-10}{15} : \dfrac{y-2}{3y}
  7. 4x+127x×5x2−10x6−9x×212x−4\dfrac{4x+12}{7x} \times \dfrac{5x^2-10x}{6-9x} \times \dfrac{21}{2x-4}

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Zgjidhja

  1. 10x(x+2)\dfrac{10}{x(x+2)}
  2. 3x×xx−1=3x−1\dfrac{3}{\cancel{x}} \times \dfrac{\cancel{x}}{x-1} = \dfrac{3}{x-1}
  3. 3x+2×3(x+2)2x+3=3x+2×3(x+2)2x+3=92x+3\dfrac{3}{x+2} \times \dfrac{3(x+2)}{2x+3} = \dfrac{3}{\cancel{x+2}} \times \dfrac{3\cancel{(x+2)}}{2x+3} = \dfrac{9}{2x+3}
  4. xx+2×x(x+2)x2=1\dfrac{x}{\cancel{x+2}} \times \dfrac{x\cancel{(x+2)}}{x^2} = 1
  5. xx+2×2x=xx+2×2x=2x+2\dfrac{x}{x+2} \times \dfrac{2}{x} = \dfrac{\cancel{x}}{x+2} \times \dfrac{2}{\cancel{x}} = \dfrac{2}{x+2}
  6. 5y−1015×3yy−2=5(y−2)15×3yy−2=5(y−2)15×3yy−2=15y15=y\dfrac{5y-10}{15} \times \dfrac{3y}{y-2} = \dfrac{5(y-2)}{15} \times \dfrac{3y}{y-2} = \dfrac{5\cancel{(y-2)}}{15} \times \dfrac{3y}{\cancel{y-2}} = \dfrac{15y}{15} = y
  7. 4(x+13)7x×5x(x−2)3(2−3x)×212(x−2)=4(x+3)7x×5x(x−2)3(2−3x)×212(x−2)=420(x+3)42(2−3x)=10(x+3)2−3x\dfrac{4(x+13)}{7x} \times \dfrac{5x(x-2)}{3(2-3x)} \times \dfrac{21}{2(x-2)} = \dfrac{4(x+3)}{7\cancel{x}} \times \dfrac{5\cancel{x}\cancel{(x-2)}}{3(2-3x)} \times \dfrac{21}{2\cancel{(x-2)}} = \dfrac{420(x+3)}{42(2-3x)} = \dfrac{10(x+3)}{2-3x}