Rumus Pangkat Rasional

RUMUS PANGKAT RASIONAL

1. a^m . aⁿ = a^{m + n}
   
2. $\frac{a^m}{a^n}=a^{m\; -\; n}$
   
3. (a^m . b^m) = (a . b)^m
   
4. $\frac{a^m}{b^m}={\left(\frac{a}{b}\right)}^m$
   
5. (a^m)ⁿ = a^mn
   
6. a⁰ = 1   , a ≠ 0
7. $a^{-m}=\frac{1}{a^m}$
   
8. $a^{\frac{m}{n}}=\sqrt[n]{a^m}$
   
9. $a^{-\frac{m}{n}}=\frac{1}{\sqrt[n]{a^m}}$
   
10. $\sqrt[m]{a}.\sqrt[m]{b}=\sqrt[m]{ab}={ab}^{\frac{1}{m}}$
11. $\frac{\sqrt[m]{a}}{\sqrt[m]{b}}=\sqrt[m]{\frac{a}{b}}={\left(\frac{a}{b}\right)}^{\frac{1}{m}}$

Contoh Soal :
Sederhanakan bentuk bilangan berpangkat berikut :

1. 2⁵ x 2²
   
2. $\frac{2^3}{2^2}$
   
3. (3²)³
   

Jawab :

1. 2⁵ x 2² = 2^{5 + 2} = 2⁷
   
2. $\frac{2^3}{2^2}=2^{3\; -\; 2}=2^1=2$
   
3. (3²)³ = 3⁶
   

Persamaan Kuadrat (Jawaban Soal Latihan 1) bagian 2

4. Selesaikan persamaan kuadrat berikut :
1. 5x² – 10x = 0
2. 2(x - 2)² = 8
3. 15x² + 16x + 4 = 0
4. $x^2-2x\sqrt{3}-1=0$
5. Selesaikan persamaan kuadrat berikut dengan cara yang paling tepat dan cepat :
1. (x - 1)(x - 2) = 12
2. (x + 1)(3x - 2) = 1 + x
3. $x+\frac{2}{x}=\frac{3+2x}{x}$
4. $\frac{x-6}{2}=-\frac{4}{x}$
5. $\sqrt{x+2}=x-4$
6. $\frac{x-\sqrt{3}}{x+\sqrt{3}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}=3$
7. $\frac{x+1}{x+2}+\frac{x+2}{x+1}=2\frac{1}{2}$
8. $x+2+\sqrt{x+2}-6=0$
9. x⁴ - 13x² + 36 = 0
10. (x² + 1)² - 3(x² + 1) + 2 = 0

Jawab :

4. 1. 5x² – 10x = 0
      5x (x - 2) = 0
      5x = 0 atau x - 2 = 0
      x = 0 atau x = 2
      
   2. 2 (x - 2)² = 8
      (x - 2)² = $\frac{8}{2}$
      (x - 2)² = 4
      x - 2 = $\sqrt{4}$
      x - 2 = ± 2
      x = 2 ± 2
      x = 2 + 2 atau x = 2 - 2
      x = 4 atau x = 0
      
   3. 15x² + 16 x + 4 = 0
      (3x + 2)(5x + 2) = 0
      3x + 2 = 0 atau 5x + 2 = 0
      3x = -2 atau 5x = -2
      x = $-\frac{2}{3}$ atau x = $-\frac{2}{5}$
      
   4. $x^2-\mathrm{2x}\sqrt{3}-1=0$
      
      $x^2-2\sqrt{3}x-1=0$
      
      $x^2-2\sqrt{3}x=1$
      
      ${\left(x-\sqrt{3}\right)}^2=1+3$
      
      ${\left(x-\sqrt{3}\right)}^2=4$
      
      $x-\sqrt{3}=\sqrt{4}$
      
      $x-\sqrt{3}=\pm 2$
      
      $x=\sqrt{3}\pm 2$
      
      $x=\sqrt{3}+2$ atau $x=\sqrt{3}-2$
      
5. 1. (x-1)(x-2) = 12
      x² - 2x - x + 2 = 12
      x² - 3x + 2 -12 = 0
      x² - 3x - 10 = 0
      (x-5)(x+2) = 0
      x - 5 = 0 atau x + 2 = 0
      x = 5              x = -2
      
   2. (x+1)(3x-2) = 1 + x
      3x² - 2x + 3x - 2 = 1 + x
      3x² + x - 2 = 1 + x
      3x² + x - x - 2 - 1 = 0
      3x² - 3 = 0
      3x² = 3
      $x^2=\frac{3}{3}$
      x² = 1
      $x=\sqrt{1}$
      x = ± 1
      x = -1 atau x = 1
      
   3. $x+\frac{2}{x}=\frac{3+2x}{x}$
      x² + 2 = 3 + 2x
      x² - 2x + 2 - 3 = 0
      x² - 2x - 1 = 0
      x² - 2x = 1
      (x - 1)² = 1 + 1
      (x - 1)² = 2
      x - 1 = $\sqrt{2}$
      x - 1 = ±$\sqrt{2}$
      x = 1 ± $\sqrt{2}$
      x = 1 + $\sqrt{2}$ atau x = 1 - $\sqrt{2}$
      
   4. $\frac{x-6}{2}=-\frac{4}{x}$
      $\frac{x}{2}-\frac{6}{2}=-\frac{4}{x}$
      $\frac{1}{2}{x}^2-\mathrm{3x}=-4$
      $\frac{1}{2}{x}^2-\mathrm{3x}+4=0$
      x² - 6x + 8 = 0
      (x - 4)(x - 2) = 0
      x - 4 = 0 atau x - 2 = 0
      x = 4 atau x = 2
      
   5. $\sqrt{x+2}=x-4$
      x + 2 = (x - 4)²
      x + 2 = (x - 4)(x - 4)
      x + 2 = x² - 4x - 4x + 16
      x + 2 = x² - 8x + 16
      x² - 8x - x + 16 - 2 = 0
      x² - 9x + 14 = 0
      (x - 7)(x - 2) = 0
      x - 7 = 0 atau x - 2 = 0
      x = 7 atau x = 2
      
   6. $\frac{x-\sqrt{3}}{x+\sqrt{3}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}=3$
      $\frac{x-\sqrt{3}}{x+\sqrt{3}}\frac{x-\sqrt{3}}{x-\sqrt{3}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}\frac{x+\sqrt{3}}{x+\sqrt{3}}=3$
      $\frac{x^2-\sqrt{3}x-\sqrt{3}x+3}{x^2-\sqrt{3}x+\sqrt{3}x-3}+\frac{x^2+\sqrt{3}x+\sqrt{3}x+3}{x^2-\sqrt{3}x+\sqrt{3}x-3}=3$
      $\frac{x^2-2\sqrt{3}x+3}{x^2-3}+\frac{x^2+2\sqrt{3}x+3}{x^2-3}=3$
      $\frac{x^2+x^2-2\sqrt{3}x+2\sqrt{3}x+3+3}{x^2-3}=3$
      $\frac{2x^2+6}{x^2-3}=3$
      2x² + 6 = 3 ( x² - 3 )
      2x² + 6 = 3x² - 9
      3x² - 2x² - 9 - 6 = 0
      x² - 15 = 0
      x² = 15
      x = $\sqrt{15}$
      x = $\pm \sqrt{15}$
      x = $\sqrt{15}$ atau x = $-\sqrt{15}$
      
   7. $\frac{x+1}{x+2}+\frac{x+2}{x+1}=2\frac{1}{2}$
      $\frac{\mathrm{(x\; +\; 1)}\mathrm{(x\; +\; 1)}}{\mathrm{(x\; +\; 2)}\mathrm{(x\; +\; 1)}}+\frac{\mathrm{(x\; +\; 2)}\mathrm{(x\; +\; 2)}}{\mathrm{(x\; +\; 1)}\mathrm{(x\; +\; 2)}}=\frac{5}{2}$
      $\frac{x^2+\mathrm{2x}+1}{x^2+\mathrm{3x}+2}+\frac{x^2+\mathrm{4x}+4}{x^2+\mathrm{3x}+2}=\frac{5}{2}$
      $\frac{x^2+x^2+\mathrm{2x}+\mathrm{4x}+1+4}{x^2+\mathrm{3x}+2}=\frac{5}{2}$
      $\frac{2x^2+\mathrm{6x}+5}{x^2+\mathrm{3x}+2}=\frac{5}{2}$
      $2x^2+\mathrm{6x}+5=\frac{5}{2}{x}^2+\mathrm{3x}+2$
      $4x^2+\mathrm{12x}+10=5x^2+\mathrm{15x}+10$
      $5x^2-4x^2+\mathrm{15x}-\mathrm{12x}+10-10=0$
      x² + 3x = 0
      x(x + 3) = 0
      x = 0 atau x + 3 = 0
                      x = 3
      
   8. $x+2+\sqrt{x+2}-6=0$
      $x+2-6+\sqrt{x+2}=0$
      $x-4+\sqrt{x+2}=0$
      $\sqrt{x+2}=\mathrm{-x}+4$
      x + 2 = (-x + 4)²
      x + 2 = (-x + 4)(-x + 4)
      x + 2 = x² - 4x - 4x + 16
      x + 2 = x² - 8x + 16
      x² - 8x - x + 16 - 2 = 0
      x² - 9x + 14 = 0
      (x - 7)(x - 2) = 0
      x - 7 = 0 atau x - 2 = 0
      x = 7 atau x = 2
      
   9. x⁴ - 13x² + 36 = 0
      (x² - 4)(x² - 9) = 0
      x² - 4 = 0 atau x² - 9 = 0
      x² = 4 atau x² = 9
      x = $\sqrt{4}$ atau x = $\sqrt{9}$
      x = ±2 atau x = ±3
      x = 2 atau x = -2 atau x = 3 atau x = -3
      
   10. (x² + 1)² - 3(x² + 1) + 2 = 0
       misal : a = x² + 1
       a² - 3a + 2 = 0
       (a - 2)(a - 1) = 0
       a - 2 = 0 atau a - 1 = 0
       a = 2 atau a = 1
       x² + 1 = 2 atau x² + 1 = 1
       x² = 2 - 1 atau x² = 1 - 1
       x² = 1 atau x² = 0
       x = $\sqrt{1}$ atau x = $\sqrt{0}$
       x = ±1 atau x = 0
       x = 1 atau x = -1 atau x = 0