Bentuk Pecahan

Faktor dari \(4x^2 + 8x + 3 = (2x + 1)(2x + 3)\)

\(\dfrac{12}{4x^2 + 8x + 3} \equiv \dfrac{A}{2x + 1} + \dfrac{B}{2x + 3}\)

\(\color{red} 2x + 1 \equiv 2x + P \rightarrow \color{blue} P = 1\)

\(\color{red} 2x + 3 \equiv 2x + Q \rightarrow \color{blue} Q = 3\)

\(\dfrac{12}{4x^2 + 8x + 3} \equiv \dfrac{A(2x + 3) + B(2x + 1)}{(2x + 1)(2x + 3)}\)

Samakan bagian pembilang,

\(12 \equiv A(2x + 3) + B(2x + 1)\)

Pilih \(x = -\dfrac{3}{2}\)

\(12 = A(0) + B(2(-\frac{3}{2}) + 1)\)

\(12 = -2B\)

\(\color{blue} B = -6\)

Pilih \(x = -\dfrac{1}{2}\)

\(12 = A(2(-\frac{1}{2}) + 3) + B(0)\)

\(12 = 2A\)

\(\color{blue} A = 6\)

\(\dfrac{1}{2}\left[ \dfrac{2}{3 \cdot 5} + \dfrac{2}{5 \cdot 7} + \dfrac{2}{7 \cdot 9} + \dotso + \dfrac{2}{2019 \cdot 2021}\right]\)

\(\dfrac{1}{2}\left[ \dfrac{1}{3}\:-\:\dfrac{1}{5} + \dfrac{1}{5} \:-\:\dfrac{1}{7} + \dfrac{1}{7} \:-\:\dfrac{1}{9} + \dotso + \dfrac{1}{2019} \:-\: \dfrac{1}{2021}\right]\)

\(\dfrac{1}{2} \left[\dfrac{1}{3} \:-\: \dfrac{1}{2021} \right]\)

\(\dfrac{1}{2} \cdot \dfrac{2018}{6063}\)

\(\dfrac{1009}{6063}\)

Faktor dari \(x^2 \:-\:4x \:-\:12\) adalah \((x\:-\:6)\) dan \((x + 2)\)

\(\dfrac{28x\:-\:24}{x^2 \:-\:4x\:-\:12} \equiv \dfrac{A}{x\:-\:6} + \dfrac{B}{x + 2}\)

\(\dfrac{28x\:-\:24}{x^2 \:-\:4x\:-\:12} \equiv \dfrac{A(x + 2) + B(x\:-\:6)}{(x\:-\:6)(x + 2)}\)

Samakan bagian pembilang,

\(28x \:-\:24 \equiv A(x + 2) + B(x\:-\:6)\)

Untuk \(x = -2\) maka \(28(-2)\:-\:24 = 0 + B(-2\:-\:6)\)

\(-80 = -8B\)

\(B = 10\)

Untuk \(x = 6\) maka \(28(6)\:-\:24 = A(6 + 2) + B(6 \:-\:6)\)

\(144 = 8A\)

\(A = 18\)

Jadi, \(\dfrac{28\:-\:24}{x^2 \:-\:4x \:-\:12} = \dfrac{18}{x\:-\:6} + \dfrac{10}{x + 2}\)

Bentuk \(\dfrac{18}{x\:-\:6}\) akan menjadi bulat jika \((x\:-\:6)\) adalah faktor-faktor dari 18.

Jadi \(x\:-\:6\) = −1, −2, −3, −6, −9, −18, 1, 2, 3, 6, 9, 18

\(x\) = 5, 4, 3, 0, −3, −12, 7, 8, 9, 12, 15, 24 … (1)

Bentuk \(\dfrac{10}{x + 2}\) akan menjadi bulat jika \(x + 2\) adalah faktor-faktor dari 10.

Jadi \(x + 2\) = −1, −2, −5, −10, 1, 2, 5, 10

\(x\) = −3, −4, −7, −12, −1, 0, 3, 8 … (2)

Himpunan bilangan bulat yang memenuhi adalah irisan dari persamaan (1) dan (2)

{−12, −3, 0, 3, 8}

Faktor dari \(16x^4 \:-\:24x^2 \:-\:27 = (4x^2\:-\:9)(4x^2 + 3)\)

\(16x^4 \:-\:24x^2 \:-\:27 = ((2x)^2\:-\:3^2)(4x^2 + 3\)

\(16x^4 \:-\:24x^2 \:-\:27 = (2x\:-\:3)(2x + 3)(4x^2 + 3)\)

\(\dfrac{12}{16x^4 \:-\:24x^2 \:-\:27} \equiv \dfrac{A}{2x\:-\:3} + \dfrac{B}{2x + 3} + \dfrac{Cx + D}{4x^2 + 3}\)

\(\color{red} 4x^2 + 3 \equiv Ex^2 + F\)

\(\color{blue} E = 4 \text{ dan } F = 3\)

\(\dfrac{12}{16x^4 \:-\:24x^2 \:-\:27} \equiv \dfrac{A(2x + 3)(4x^2 + 3) + B(2x\:-\:3)(4x^2 + 3) + (Cx + D)(2x\:-\:3)(2x + 3)}{(2x\:-\:3)(2x + 3)(4x^2 + 3)}\)

Samakan bagian pembilang,

\(12 \equiv A(2x + 3)(4x^2 + 3) + B(2x\:-\:3)(4x^2 + 3) + (Cx + D)(2x\:-\:3)(2x + 3)\)

Pilih \(x = -\dfrac{3}{2}\) akan diperoleh nilai \(\color{blue} B = -\dfrac{1}{6}\)

Pilih \(x = \dfrac{3}{2}\) akan diperoleh nilai \(\color{blue} A = \dfrac{1}{6}\)

Pilih \(x = 0\) akan diperoleh nilai \(\color{blue} D = -1\)

Pilih \(x = 1\) akan diperoleh nilai \(\color{blue} C = 0\)