Nyatakan fungsi pecahan berikut menjadi fungsi pecahan sebagian. (11x² + 4x + 12)/[(2x + 1)(x² + 4)]

<p>Jawaban : &nbsp;(3/(2x + 1)) + ((4x)/(x² + 4)).</p><p>&nbsp;</p><p>Ingat!<br>Jika (ax²+bx+c) /d = (ex²+fx+g)/d<br>Maka a=e, b=f, c=g.</p><p>(a+b) (c+d) = ac +ad + bc + bd.</p><p>&nbsp;</p><p>Perhatikan.</p><p>(11x² + 4x + 12)/[(2x + 1)(x² + 4)] ≡ (A/(2x + 1)) + ((Bx+C)/(x² + 4))</p><p>(11x² + 4x + 12)/[(2x + 1)(x² + 4)] ≡ (A(x² + 4)/[(2x + 1)(x² + 4)]) + ((Bx+C)(2x + 1)/[(2x + 1)(x² + 4)]</p><p>Sehingga</p><p>11x² + 4x + 12 = A(x² + 4)+(Bx+C)(2x + 1)</p><p>Untuk x = -1/2 maka</p><p>11.(-1/2)² + 4.(-1/2) + 12 = A((-1/2)² + 4)+(B.(-1/2)+C)(2.(-1/2) + 1)</p><p>11.(1/4) - 2 + 12 = A((1/4) + 4)+(B.(-1/2)+C)(-1 + 1)</p><p>(11/4) + 10 = A((1/4) + (16/4))+0</p><p>(11/4) + (40/4) = A(17/4)</p><p>(51/4) = A(17/4)</p><p>A = (51/4) . (4/17)</p><p>A = 51/17</p><p>A = 3</p><p>&nbsp;</p><p>Perhatikan.</p><p>11x² + 4x + 12 = A(x² + 4)+(Bx+C)(2x + 1)</p><p>11x² + 4x + 12 = Ax² + 4A+2Bx²+Bx+2Cx+C</p><p>11x² + 4x + 12 = Ax²+2Bx²+Bx+2Cx + 4A+C</p><p>11x² + 4x + 12 = (A+2B)x²+(B+2C)x + (4A+C)</p><p>&nbsp;</p><p>Diperoleh</p><p>A+2B = 11 .....(1)</p><p>B+2C = 4 .....(2)</p><p>4A+C = 12 .....(3)</p><p>&nbsp;</p><p>Substitusi A = 3 ke A+2B = 11.<br>3+2B = 11</p><p>2B = 11=3</p><p>2B = 8</p><p>B = 8/2</p><p>B = 4</p><p>&nbsp;</p><p>Substitusi B = 4 ke B+2C = 4.</p><p>4+2C = 4</p><p>2C = 4-4</p><p>2C = 0</p><p>C = 0/2</p><p>C = 0</p><p>&nbsp;</p><p>Nilai A = 3, B = 4 dan C = 0, sehingga&nbsp;</p><p>(11x² + 4x + 12)/[(2x + 1)(x² + 4)] ≡ (3/(2x + 1)) + ((4x+0)/(x² + 4))</p><p>(11x² + 4x + 12)/[(2x + 1)(x² + 4)] ≡ (3/(2x + 1)) + ((4x)/(x² + 4)).</p><p>&nbsp;</p><p>Jadi, jawaban yang benar adalah (3/(2x + 1)) + ((4x)/(x² + 4)).</p>