Resha A
27 Februari 2023 03:36
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Resha A
27 Februari 2023 03:36
Pertanyaan
Nyatakan setiap fungsi pecahan berikut menjadi fungsi pecahan sebagian. d. (x^2 - 7)/(3x^3 + x^2 + x - 2)
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L. Nikmah
Mahasiswa/Alumni Universitas Pendidikan Indonesia
09 Maret 2023 01:30
<p>Jawaban : (-59)/[75x-50)] + (28x+58)/[25x<sup>2 </sup>+25 x + 25].</p><p> </p><p>Konsep!<br>(ax + b)/(x+p)(x+q) = A/(x+p) + B/(x+q)</p><p> </p><p>Perhatikan metode horner di bawah ini.</p><p>(x<sup>2 </sup>- 7)/(3x<sup>3</sup> + x<sup>2 </sup> + x - 2) = (x<sup>2</sup> - 7)/[(3x-2)(x<sup>2 </sup>+ x + 1)<br><br>Fungsi pecahan sebagian.</p><p>(x<sup>2 </sup>- 7)/(3x<sup>3</sup> + x<sup>2 </sup> + x - 2) = A/(3x-2) + (Bx+C)/(x<sup>2 </sup>+ x + 1)</p><p>(x<sup>2 </sup>- 7)/(3x<sup>3</sup> + x<sup>2 </sup> + x - 2) = [A(x<sup>2 </sup>+ x + 1)]/[(3x-2)(x<sup>2 </sup>+ x + 1)] + [(Bx+C)(3x-2)]/[(3x-2)(x<sup>2 </sup>+ x + 1)]</p><p>Diperoleh</p><p>(x<sup>2 </sup>- 7) = [A(x<sup>2 </sup>+ x + 1)] + [(Bx+C)(3x-2)]</p><p>Untuk x = 2/3 maka</p><p>((2/3)<sup>2 </sup>- 7) = [A((2/3)<sup>2 </sup>+ (2/3) + 1)] + [(B(2/3)+C)(3(2/3)-2)]</p><p>((4/9)<sup> </sup>- 7) = [A((4/9)<sup> </sup>+ (2/3) + 1)] + 0</p><p>(4/9<sup> </sup>- 63/9) = [A(4/9<sup> </sup>+ 12/9 + 9/9)]</p><p>(-59/9) = A(25/9)</p><p>A = (-59/9).(9/25)</p><p>A = (-59/25)</p><p>Menentukan B dan C.</p><p>(x<sup>2 </sup>- 7) = [Ax<sup>2 </sup>+ Ax + A] + [3Bx<sup>2</sup>-2Bx+3Cx-2C)]</p><p>(x<sup>2 </sup>- 7) = [Ax<sup>2 </sup>+ Ax + A + 3Bx<sup>2</sup>-2Bx+3Cx-2C]</p><p>(x<sup>2 </sup>- 7) = [Ax<sup>2 </sup> + 3Bx<sup>2 </sup>+ Ax-2Bx+3Cx + A-2C]</p><p>(x<sup>2 </sup>- 7) = [(A+3B)x<sup>2 </sup>+ (A-2B+3C)x + (A-2C)]</p><p> </p><p>Diperoleh</p><p>1 = A+3B</p><p>1 = (-59/25)+3B</p><p>3B = 1+(59/25)</p><p>3B = (25/25)+(59/25)</p><p>3B = (84/25)</p><p>B = (84/25).(1/3)</p><p>B = (28/25)</p><p> </p><p>A-2C = -7</p><p>(-59/25)-2C = -7</p><p>2C = (-59/25)+7</p><p>2C = (-59/25)+(175/25)</p><p>2C = (116/25)</p><p>C = (116/25).(1/2)</p><p>C = (58/25)</p><p>Sehingga, </p><p>(x<sup>2 </sup>- 7)/(3x<sup>3</sup> + x<sup>2 </sup> + x - 2) = A/(3x-2) + (Bx+C)/(x<sup>2 </sup>+ x + 1)</p><p>(x<sup>2 </sup>- 7)/(3x<sup>3</sup> + x<sup>2 </sup> + x - 2) = (-59/25)/(3x-2) + ((28/25)x+(58/25))/(x<sup>2 </sup>+ x + 1)</p><p>(x<sup>2 </sup>- 7)/(3x<sup>3</sup> + x<sup>2 </sup> + x - 2) = (-59)/[75x-50)] + (28x+58)/[25x<sup>2 </sup>+25 x + 25]</p><p> </p><p>Jadi, (x<sup>2 </sup>- 7)/(3x<sup>3</sup> + x<sup>2 </sup> + x - 2) = (-59)/[75x-50)] + (28x+58)/[25x<sup>2 </sup>+25 x + 25].</p>
Jawaban : (-59)/[75x-50)] + (28x+58)/[25x2 +25 x + 25].
Konsep!
(ax + b)/(x+p)(x+q) = A/(x+p) + B/(x+q)
Perhatikan metode horner di bawah ini.
(x2 - 7)/(3x3 + x2 + x - 2) = (x2 - 7)/[(3x-2)(x2 + x + 1)
Fungsi pecahan sebagian.
(x2 - 7)/(3x3 + x2 + x - 2) = A/(3x-2) + (Bx+C)/(x2 + x + 1)
(x2 - 7)/(3x3 + x2 + x - 2) = [A(x2 + x + 1)]/[(3x-2)(x2 + x + 1)] + [(Bx+C)(3x-2)]/[(3x-2)(x2 + x + 1)]
Diperoleh
(x2 - 7) = [A(x2 + x + 1)] + [(Bx+C)(3x-2)]
Untuk x = 2/3 maka
((2/3)2 - 7) = [A((2/3)2 + (2/3) + 1)] + [(B(2/3)+C)(3(2/3)-2)]
((4/9) - 7) = [A((4/9) + (2/3) + 1)] + 0
(4/9 - 63/9) = [A(4/9 + 12/9 + 9/9)]
(-59/9) = A(25/9)
A = (-59/9).(9/25)
A = (-59/25)
Menentukan B dan C.
(x2 - 7) = [Ax2 + Ax + A] + [3Bx2-2Bx+3Cx-2C)]
(x2 - 7) = [Ax2 + Ax + A + 3Bx2-2Bx+3Cx-2C]
(x2 - 7) = [Ax2 + 3Bx2 + Ax-2Bx+3Cx + A-2C]
(x2 - 7) = [(A+3B)x2 + (A-2B+3C)x + (A-2C)]
Diperoleh
1 = A+3B
1 = (-59/25)+3B
3B = 1+(59/25)
3B = (25/25)+(59/25)
3B = (84/25)
B = (84/25).(1/3)
B = (28/25)
A-2C = -7
(-59/25)-2C = -7
2C = (-59/25)+7
2C = (-59/25)+(175/25)
2C = (116/25)
C = (116/25).(1/2)
C = (58/25)
Sehingga,
(x2 - 7)/(3x3 + x2 + x - 2) = A/(3x-2) + (Bx+C)/(x2 + x + 1)
(x2 - 7)/(3x3 + x2 + x - 2) = (-59/25)/(3x-2) + ((28/25)x+(58/25))/(x2 + x + 1)
(x2 - 7)/(3x3 + x2 + x - 2) = (-59)/[75x-50)] + (28x+58)/[25x2 +25 x + 25]
Jadi, (x2 - 7)/(3x3 + x2 + x - 2) = (-59)/[75x-50)] + (28x+58)/[25x2 +25 x + 25].
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