would someone help with this math problem?
回答 (5)
2009-07-02 6:22 am
✔ 最佳答案
(x^3 - 27) / (3x^2-8x-3)= [(x-3)(x^2+3x+9)] / [(3x+1)(x-3)]
= (x^2+3x+9) / (3x+1)
2009-07-02 1:34 pm
= (x³ - 27)/(3x² - 8x - 3)
= ([x - 3][x² + 3x - 9])/([x - 3][3x + 1])
= (x² + 3x - 9)/(3x + 1)
Answer: (x² + 3x - 9)/(3x + 1)
= ([x - 3][x² + 3x - 9])/([x - 3][3x + 1])
= (x² + 3x - 9)/(3x + 1)
Answer: (x² + 3x - 9)/(3x + 1)
2009-07-02 3:37 pm
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
(x^3 - 27)/(3x^2 - 8x - 3)
= (x^3 - 3^3)/(3x^2 + x - 9x - 3)
= [(x - 3)(x^2 + 3x + 9)]/[(3x^2 + x) - (9x + 3)]
= [(x - 3)(x^2 + 3x + 9)]/[x(3x + 1) - 3(3x + 1)]
= [(x - 3)(x^2 + 3x + 9)]/[(3x + 1)(x - 3)]
= (x^2 + 3x + 9)/(3x + 1)
(x^3 - 27)/(3x^2 - 8x - 3)
= (x^3 - 3^3)/(3x^2 + x - 9x - 3)
= [(x - 3)(x^2 + 3x + 9)]/[(3x^2 + x) - (9x + 3)]
= [(x - 3)(x^2 + 3x + 9)]/[x(3x + 1) - 3(3x + 1)]
= [(x - 3)(x^2 + 3x + 9)]/[(3x + 1)(x - 3)]
= (x^2 + 3x + 9)/(3x + 1)
2009-07-02 1:26 pm
(x-3)(x^2-3x+9)/(3x-5)(x+1/6)
2009-07-02 1:26 pm
I confirm peter's answer. I got the same thing he did.
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