Toán Lớp 8: Giúp Mình Nhoa !!( > ω < ) Làm tính nhân: $\frac{x^3-1}{2x^2+13x-24}$ . $\frac{6x^3+39x^2-72x}{x^3-x^2+3x-3}$

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1. Gửi bạn:

   $ĐKXĐ:x \neq 0;1$

   $\dfrac{x^3-1}{2x^2+13x-24}.\dfrac{6x^3+39x^2-72x}{x^3-x^2+3x-3}$ 

   $=\dfrac{x^3-1}{2x^2+13x-24}.\dfrac{3x(2x^2+13x-24)}{x^3-x^2+3x-3}$ 

   $=\dfrac{(x-1)(x^2+x+1)}{2x^2+13x-24}.\dfrac{3x(2x^2+13x-24)}{x^2(x-1)+3(x-1)}$ 

   $=\dfrac{(x-1)(x^2+x+1)}{2x^2+13x-24}.\dfrac{3x(2x^2+13x-24)}{(x^2+3)(x-1)}$ 

   $=\dfrac{x^2+x+1}{1}.\dfrac{3x}{x^2+3}$ 

   $=\dfrac{3x(x^2+x+1)}{x^2+3}$ 

    

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