Question

решите уравнение 7x-6/x 3+27 1/x 2-3x+9-1/x+3 даю 20 баллов​

Answer 1

$\displaystyle\bf\\\frac{7x-6}{x^{3}+27 } =\frac{1}{x^{2} -3x+9} -\frac{1}{x+3} \\\\\\\frac{7x-6}{(x+3)(x^{2} -3x+9)} -\frac{1}{x^{2} -3x+9} +\frac{1}{x+3} =0\\\\\\\frac{7x-6-(x+3)+x^{2} -3x+9}{(x+3)(x^{2} -3x+9)}=0\\\\\\\frac{7x-6-x-3+x^{2} -3x+9}{(x+3)(x^{2} -3x+9)}=0 \\\\\\\frac{x^{2} +3x}{(x+3)(x^{2} -3x+9)}=0 \\\\\\\frac{x(x +3)}{(x+3)(x^{2} -3x+9)}=0\\\\\\x=0 \ ; \ x+3\neq 0\\\\\\Otvet:0$