Question

объясните как это решать, пожалуйста!
$\frac{7x-6}{x^{3}+27 } =\frac{1}{x^{2}-3x+9 } -\frac{1}{x+3}$

Answer 1

$\displaystyle\bf\\\frac{7x-6}{x^{3}+27 } =\frac{1}{x^{2} -3x+9}-\frac{1}{x+3} \\\\\\\frac{7x-6}{(x+3)(x^{2} -3x+9)} -\frac{1}{x^{2} -3x+9} +\frac{1}{x+3} =0\\\\\\\frac{7x-6-1\cdot(x+3)+1\cdot(x^{2} -3x+9)}{(x+3)(x^{2} -3x+9)} =0\\\\\\\frac{7x-6-x-3+x^{2} -3x+9}{(x+3)(x^{2} -3x+9)} =0\\\\\\\frac{x^{2} +3x}{(x+3)(x^{2} -3x+9)} =0\\\\\\\left \{ {{x^{2} +3x=0} \atop {x+3\neq 0 \ \ , \ \ x^{2} -3x+9\neq 0}} \right.$

$\displaystyle\bf\\x^{2} +3x=0 \ \ ; \ x\neq -3\\\\x(x+3)=0\\\\x=0 \ \ x+3\neq 0\\\\Otvet: \ 0$