Giải thích các bước giải:

7.ĐKXĐ: $x\ne 3$

Ta có:

$\dfrac{x^2-3}{x^3-27}=\dfrac1{x-3}-\dfrac3{x^2+3x+4}$ 

$\to \dfrac{x^2-3}{(x-3)(x^2+3x+9)}=\dfrac1{x-3}-\dfrac3{x^2+3x+4}$ 

$\to \left(x^2-3\right)\left(x^2+3x+4\right)=\left(x^2+3x+9\right)\left(x^2+3x+4\right)-3\left(x-3\right)\left(x^2+3x+9\right)$

$\to x^4+3x^3+x^2-9x-12=x^4+3x^3+22x^2+39x+117$

$\to 21x^2+48x+129=0$

$\to 7x^2+16x+43=0$

$\to 7\left(x+\dfrac{8}{7}\right)^2+\dfrac{237}{7}=0$ vô nghiệm vì $7\left(x+\dfrac{8}{7}\right)^2+\dfrac{237}{7}>0$

8.ĐKXĐ: $x\ne 3, x\ne 4$

Ta có:

$\dfrac{12}{x-4}=\dfrac3{x-3}-\dfrac{3x+7}{x^2-7x+12}$

$\to \dfrac{12}{x-4}=\dfrac3{x-3}-\dfrac{3x+7}{(x-3)(x-4)}$

$\to 12(x-3)=3(x-4)-(3x+7)$

$\to 12x-36=-19$

$\to 12x=17$

$\to x=\dfrac{17}{12}$