Treść zadania

Rozwiązania

• 

  1 0

  antekL1 1.3.2013 (19:55)

  Zadanie 1.
  a)
  \left(3x-\frac{1}{3}\right)\left(2x-\frac{1}{2}\right)=6x^2-\frac{3}{2}x-\frac{2}{3}x+\frac{1}{6} = 6x^2-2\frac{1}{6}x+\frac{1}{6}
  b)
  (1{,}2a+0{,}4b)(0{,}4a+1{,}2b)=0{,}48a^2+1{,}44ab+0{,}16ab+0{,}48b^2=
  
  =0{,}48a^2+1{,}6ab+0{,}48b^2
  c)
  (x^3+x^2+ x + 1)(x-1) = x^4+x^3+x^2 + x - x^3-x^2-x - 1 = x^4-1
  =============================
  
  Zadanie 2.
  a)
  \left(16\cdot 2^3\right)^4=\left(2^4\cdot 2^3\right)^4=2^{(4+3)\cdot 4}=2^{28}
  b)
  27^2\cdot 9^6=\left(3^3\right)^2\cdot \left(3^2\right)^6=3^{3\cdot 2+ 2\cdot 6} = 3^{18}
  =============================
  
  Zadanie 3.
  a)
  =\frac{\left(-\frac{3}{2}\right)^3\cdot\left(-\frac{4}{3}\right)^3\cdot(-6)^3}{\left(-\frac{4}{3}\right)^3\cdot(-3)^3} = \left[\frac{\left(-\frac{3}{2}\right)\cdot\left(-\frac{4}{3}\right)\cdot(-6)}{\left(-\frac{4}{3}\right)\cdot(-3)}\right]^3 = (-3)^3=27
  b)
  \frac{5^8\cdot2^9}{10^7}=\frac{5^8\cdot2^9}{5^7\cdot 2^7}=5^1\cdot 2^2=5\cdot4 = 20
  =============================
  
  Zadanie 4.
  a)
  \sqrt{128} = \sqrt{2^6\cdot 2} = 2^3\sqrt{2} = 8\sqrt{2}
  b)
  \sqrt[3]{2000}=\sqrt[3]{10^3\cdot 2}=10\sqrt[3]{2}
  =============================
  
  Zadanie 5.
  a)
  \sqrt{2}\left(\sqrt{18}+\sqrt{50}\,\right)=\sqrt{36}+\sqrt{100}=6+10=16
  b)
  =\frac{\sqrt[3]{0{,}02}}{\sqrt[3]{0{,}16}}-\frac{\sqrt{\frac{3}{20}}}{\sqrt{\frac{3}{5}}}=\sqrt[3]{0{,}125}-\sqrt{\frac{1}{4}}=0{,}5-\frac{1}{2} = 0
  =============================

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