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\(\left(\frac{x^2+3x}{x^3+3x^2+9x+27}\right)\): \(\left(\frac{1}{x-3}-\frac{6x}{x^3-3x^2+9x-27}\right)\)
=\(\left[\frac{x\left(x+3\right)}{x^2\left(x+3\right)+9\left(x+3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{x^2\left(x-3\right)+9\left(x-3\right)}\right]\)
=\(\left[\frac{x\left(x-3\right)}{\left(x^2+9\right)\left(x-3\right)}\right]\):\(\left[\frac{1}{x-3}-\frac{6x}{\left(x^2+9\right)\left(x-3\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\left[\frac{x^2+9}{\left(x-3\right)\left(x^2+9\right)}-\frac{6x}{\left(x-3\right)\left(x^2+9\right)}\right]\)
=\(\frac{x}{x^2+9}\):\(\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x^2+9\right)}\)
=\(\frac{x}{x^2+9}\):\(\frac{x-3}{x^2+9}\)
=\(\frac{x}{x^2+9}\).\(\frac{x^2+9}{x-3}\)
=\(\frac{x}{x-3}\)
ĐKXĐ : \(\hept{\begin{cases}x-2\ne0\\3-4x\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne2\\x\ne\frac{3}{4}\end{cases}}}\)
\(\frac{5}{x-2}+\frac{6}{3-4x}=0\)
\(\frac{5\left(3-4x\right)}{\left(x-2\right)\left(3-4x\right)}+\frac{6\left(x-2\right)}{\left(3-4x\right)\left(x-2\right)}=0\)
\(15-20x+6x-12=0\)
\(3-14x=0\Leftrightarrow14x=3\Leftrightarrow x=\frac{3}{14}\)theo ĐKXĐ : x thỏa mãn