Đáp án:

Giải thích các bước giải:

 $a, \dfrac{3x^{2}-12x+12}{x^{4}-8x}=\dfrac{3(x^{2}-4x+4)}{x(x^{3}-8)}=\dfrac{3(x-2)^{2}}{x(x-2)(x^{2}+2x+4)}=\dfrac{3(x-2)}{x(x^{2}+2x+4)}$

$b,\dfrac{7x^{2}+14x+7}{3x^{2}+3x}=\dfrac{7(x^{2}+2x+1)}{3x(x+1)}=\dfrac{7(x+1)^{2}}{3x(x+1)}=\dfrac{7(x+1)}{3x}$

(bài tham khảo)

$\\a,\\\dfrac{3x^2-12+12}{x^4-8x}(x\ne 0, x\ne 2)\\=\dfrac{3(x^2-4x+4)}{x (x^3-8)}\\=\dfrac{3(x^2-2.x.2+2^2)}{x(x-2)(x^2+2x+4)}\\=\dfrac{3(x-2)^2}{x(x-2)(x^2+2x+4)}\\=\dfrac{3(x-2)}{x(x^2+2x+4)}\\=\dfrac{3x-6}{x^3+2x^2+4x}\\b,\\\dfrac{7x^2+14x+7}{3x^2+3x}(x\ne 0,x\ne -1)\\=\dfrac{7(x^2+2.x.1+1^2)}{3x(x+1)}\\=\dfrac{7(x+1)^2}{3x(x+1)}\\=\dfrac{7(x+1)}{3x}\\=\dfrac{7x+7}{3x}$