Matematika

Pertanyaan

sederhanakan bentuk pecahan berikut !! :) THANKS
1. \frac{3c}{6ac + 9bc} ?
2. \frac{x^{2} + xy - 12y^{2} }{ x^{2} - 16y^{2} } ?
3. \frac{4 m^{2} - 9}{3 + 2m} ?
4. \frac{5}{x + y} - \frac{2}{x - y} ?
5. \frac{3}{a + 1} - \frac{1}{a^{2} - a} ??

1 Jawaban

  • 1) [tex] \frac{3c}{6ac + 9bc} [/tex] = [tex] \frac{3c}{3c(2a + 3b)} [/tex] = [tex] \frac{1}{2a + 3b} [/tex]

    2) [tex] \frac{x^{2} + xy - 12y^{2} }{ x^{2} - 16y^{2} } [/tex] = [tex] \frac{(x + 4y) (x - 3y)}{(x - 4y) (x + 4y)} [/tex] = [tex] \frac{x - 3y}{x - 4y} [/tex]

    3) [tex] \frac{4 m^{2} - 9}{3 + 2m} [/tex] = [tex] \frac{(2m - 3) (2m + 3)}{3 + 2m} [/tex] = 2m - 3

    4) [tex] \frac{5}{x + y} - \frac{2}{x - y} [/tex] = [tex] \frac{5(x - y) - 2(x + y)}{(x + y)(x - y)} [/tex] = [tex] \frac{5x - 5y - 2x - 2y}{x^{2} - y^{2}} [/tex] = [tex] \frac{3x - 7y}{x^{2} - y^{2}} [/tex]

    5) [tex] \frac{3}{a + 1} - \frac{1}{a^{2} - a} [/tex] = [tex] \frac{3 (a^{2} - a) - (a + 1)}{(a + 1)(a^{2} - a)} [/tex] = [tex] \frac{3a^{2} - 4a - 1}{a (a - 1)(a + 1)} [/tex] = [tex] \frac{(3a - 1)(a - 1)}{a (a - 1)(a + 1)} [/tex] = [tex] \frac{3a - 1}{a^{2} + a} [/tex]

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