Pangkat Rasional (Soal Jawab)

Soal

1. Sederhanakan bentuk bilangan berpangkat berikut :
   
   1. 2² x 2³
      
   2. (3²)⁵ . 3⁵
      
   3. ${\left(\frac{2}{3}\right)}^2.{\left(\frac{3}{2}\right)}^3$
      
   4. (0,125)² x (0,125)³
      
   5. (2⁴)² : 2³
      
   6. (5⁴)³ x 5⁵
      
   7. (3⁴ x 3³) x 3⁵
      
   8. (4⁶ + 8³) 2³
      
   9. (81³ - 243) + 16²
      
   10. (6² x 6³) x 6⁴
       
   11. ${\left(\frac{3}{4}\right)}^{4}x{\left(\frac{2}{3}\right)}^3$
       
2. Hitung hasilnya dengan mengubah menjadi bilangan berpangkat terlebih dahulu !
   1. $\sqrt[3]{\frac{1}{9}}x\sqrt[3]{\frac{1}{3}}x\frac{1}{3}\sqrt[4]{9}x9\sqrt[6]{8}$
   2. $(8^{-\frac{1}{3}}+8^{-2}-8^{-\frac{2}{3}}+8^{\frac{2}{3}}):\left(2^{-2}\right)$
   3. $\frac{{(16)}^{-\frac{3}{4}}+{(8)}^{-\frac{4}{3}}}{\mathrm{(2)}{(4)}^{-\frac{3}{2}}+8^{-1}}$
   4. $\frac{(a^{\frac{3}{2}}+a^{\frac{1}{2}})}{(a^{\frac{1}{2}}+a^{-\frac{1}{2}})}x\frac{(a^{\frac{1}{3}}+a^{-\frac{1}{3}})}{(\mathrm{a}+a^{\frac{1}{3}})}$

Jawab :

1. 1. 2² x 2³ = 2^{2 + 3} = 2⁵
      
   2. (3²)⁵ . 3⁵ = 3^{2 x 5} . 3⁵ = 3¹⁰ . 3⁵ = 3^{10 + 5} = 3¹⁵
      
   3. ${\left(\frac{2}{3}\right)}^2.{\left(\frac{3}{2}\right)}^3=\frac{2^2}{3^2}.\frac{3^3}{2^3}=\frac{2^2}{2^3}.\frac{3^3}{3^2}$ = 2^{2 - 3} . 3^{3 - 2} = 2^-1 . 3¹ = $\frac{3}{2}$
      
   4. (0,125)² x (0,125)³ = (0,125)^{2 + 3} = (0,125)⁵ = ${\left(\frac{1}{8}\right)}^5={\left(\frac{1}{2^3}\right)}^5$ = (2^-3)⁵ = 2^-15
   5. (2⁴)² : 2³ = 2^{4 x 2} : 2³ = 2⁸ : 2³ = 2^{8 - 3} = 2⁵
   6. (5⁴)³ x 5⁵ = 5^{4 x 3} x 5⁵ = 5¹² x 5⁵ = 5^{12 + 5} = 5¹⁷
   7. (3⁴ x 3³) x 3⁵ = (3^{4 + 3}) x 3⁵ = 3⁷ x 3⁵ = 3^{7 + 5} = 3¹²
   8. (4⁶ + 8³) 2³ = ((2²)⁶ + (2³)³) 2³ = (2¹² + 2⁹) 2³ = 2^{12 + 3} + 2^{9 + 3} = 2¹⁵ + 2¹²
   9. (81³ - 243) + 16² = ((3⁴)³ - 3⁵) + (2⁴)² = 3¹² - 3⁵ + 2⁸
   10. (6² x 6³) x 6⁴ = 6^{2 + 3} x 6⁴ = 6⁵ x 6⁴ = 6^{5 + 4} = 6⁹
   11. ${\left(\frac{3}{4}\right)}^{4}x{\left(\frac{2}{3}\right)}^3=\frac{3^4}{{\left(2^2\right)}^4}x\frac{2^3}{3^3}=\frac{3^4}{2^8}x\frac{2^3}{3^3}=\frac{3^4}{3^3}x\frac{2^3}{2^8}=3^{4\; -\; 3}x{2}^{3\; -\; 8}=3^{1}x{2}^{-5}=\frac{3}{2^5}$
       
2. 1. $\sqrt[3]{\frac{1}{9}}x\sqrt[3]{\frac{1}{3}}x\frac{1}{3}\sqrt[4]{9}x9\sqrt[6]{8}$
      
      = $3^{-\frac{2}{3}}x{3}^{-\frac{1}{3}}x{3}^{-1}{3}^{\frac{2}{4}}x{3}^2{2}^{\frac{3}{6}}$
      
      = $3^{-\frac{2}{3}-\frac{1}{3}-1+\frac{1}{2}}x{3}^2{2}^{\frac{1}{2}}$
      
      = $3^{-\frac{3}{2}+2}x{2}^{\frac{1}{2}}$
      
      = $3^{\frac{1}{2}}{2}^{\frac{1}{2}}$
      
      = $\sqrt{3}\sqrt{2}$
   2. $(8^{-\frac{1}{3}}+8^{-2}-8^{-\frac{2}{3}}+8^{\frac{2}{3}}):\left(2^{-2}\right)$
      
      = $({\left(2^3\right)}^{-\frac{1}{3}}+{\left(2^3\right)}^{-2}-{\left(2^3\right)}^{-\frac{2}{3}}+{\left(2^3\right)}^{\frac{2}{3}}):\left(2^{-2}\right)$
      
      = $(2^{-1}+2^{-6}-2^{-2}+2^2):\left(2^{-2}\right)$
      
      = $\left((,2^{-1},:,2^{-2},),+,(,2^{-6},:,2^{-2},),-,(,2^{-2},:,2^{-2},),+,(,2^2,:,2^{-2},)\right)$
      
      = $\left(2^{-1-(-2)},+,2^{-6-(-2)},-,2^{-2-(-2)},+,2^{2-(-2)}\right)$
      
      = $\left(2^1,+,2^{-4},-,2^0,+,2^4\right)$
      
      = $\left(2,+,2^{-4},-,1,+,2^4\right)$
      
      = $\left(1,+,2^{-4},+,2^4\right)$
      
      
   3. $\frac{{(16)}^{-\frac{3}{4}}+{(8)}^{-\frac{4}{3}}}{\mathrm{(2)}{(4)}^{-\frac{3}{2}}+8^{-1}}$
      
      = $\frac{{\left(2^4\right)}^{-\frac{3}{4}}+{\left(2^3\right)}^{-\frac{4}{3}}}{\mathrm{(2)}{\left(2^2\right)}^{-\frac{3}{2}}+{\left(2^3\right)}^{-1}}$
      
      = $\frac{2^{-3}+2^{-4}}{\mathrm{(2)}{2}^{-3}+2^{-3}}$
      
      = $\frac{\frac{1}{2^3}+\frac{1}{2^4}}{\frac{2}{2^3}+\frac{1}{2^3}}$
      
      = $\frac{\frac{2}{2^4}+\frac{1}{2^4}}{\frac{2}{2^3}+\frac{1}{2^3}}$
      
      = $\frac{\frac{3}{2^4}}{\frac{3}{2^3}}$
      
      = $\frac{3}{2^4}x\frac{2^3}{3}$
      
      = $\frac{1}{2}$
      
      
   4. $\frac{(a^{\frac{3}{2}}+a^{\frac{1}{2}})}{(a^{\frac{1}{2}}+a^{-\frac{1}{2}})}x\frac{(a^{\frac{1}{3}}+a^{-\frac{1}{3}})}{(\mathrm{a}+a^{\frac{1}{3}})}$
      
      = $\frac{(\mathrm{a\; .}{a}^{\frac{1}{2}}+a^{\frac{1}{2}})}{(a^{\frac{1}{2}}+\frac{1}{a^{\frac{1}{2}}})}x\frac{(a^{\frac{1}{3}}+\frac{1}{a^{\frac{1}{3}}})}{(\mathrm{a}+a^{\frac{1}{3}})}$
      
      = $\frac{\mathrm{a}(a^{\frac{1}{2}}+\frac{1}{a^{\frac{1}{2}}})}{(a^{\frac{1}{2}}+\frac{1}{a^{\frac{1}{2}}})}x\frac{a^{\frac{1}{3}}(1+\frac{1}{a^{\frac{2}{3}}})}{\mathrm{a}(1+\frac{1}{a^{\frac{2}{3}}})}$
      
      = $\mathrm{a}x\frac{a^{\frac{1}{3}}}{\mathrm{a}}$
      
      = $a^{\frac{1}{3}}$