Kopertina e librit Matematika 10 - 11: Pjesa I

Zgjidhja e ushtrimit 15

Zgjidhja e ushtrimit 15 të mësimit 2.2A 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. (8x2×2x124x3)3(\dfrac{8x^2 \times \sqrt{2x^{\frac{1}{2}}}}{4x^3})^3
  2. (y−2×2y432y23)45(\dfrac{y^{-2} \times 2y^4}{32y^{\frac{2}{3}}})^{\frac{4}{5}}

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Zgjidhja

  1. (8x2×(2x12)124x3)3=(8x2×2x14)4x3)3=(82x2+14)4x3)3=(82x94)4x3)3=(22x94−3)3=(22x−34)3=23×(2)3×x−34×3=162x−94\begin{aligned} (\frac{8x^2 \times (2x^{\frac{1}{2}})^{\frac{1}{2}}}{4x^3})^3&=(\frac{8x^2 \times \sqrt{2}x^{\frac{1}{4}})}{4x^3})^3 \\ &=(\frac{8 \sqrt{2}x^{2+\frac{1}{4}})}{4x^3})^3 \\ &=(\frac{8 \sqrt{2}x^{\frac{9}{4}})}{4x^3})^3 \\ &= (2\sqrt{2}x^{\frac{9}{4} - 3})^3 \\ &= (2\sqrt{2}x^{-\frac{3}{4}})^3 \\ &= 2^3\times (\sqrt{2})^3 \times x^{-\frac{3}{4} \times3} \\ &= 16\sqrt{2}x^{-\frac{9}{4}} \end{aligned}
  2. (2y−2+432y23)45=(2y232y23)45=(116y2−23)45=(116y43)45=(116)45y43×45=(116)45y1615\begin{aligned} (\frac{2y^{-2 + 4}}{32y^{\frac{2}{3}}})^{\frac{4}{5}}&=(\frac{2y^{2}}{32y^{\frac{2}{3}}})^{\frac{4}{5}} \\ &=(\frac{1}{16}y^{2-\frac{2}{3}})^{\frac{4}{5}} \\ &=(\frac{1}{16}y^{\frac{4}{3}})^{\frac{4}{5}} \\ &=(\frac{1}{16})^{\frac{4}{5}}y^{\frac{4}{3} \times \frac{4}{5}} \\ &=(\frac{1}{16})^{\frac{4}{5}}y^{\frac{16}{15}} \\ \end{aligned}