Question

С розв'язанням даю 100 балов

Answer 1

$\displaystyle\bf\\\frac{6x+14}{x^{2} -9} +\frac{7}{x^{2} +3x} =\frac{6}{x-3} \\\\\\\frac{6x+14}{(x -3)(x+3)} +\frac{7}{x\cdot(x +3)} -\frac{6}{x-3} =0\\\\\\\frac{(6x+14)\cdot x+7\cdot(x-3)-6\cdot(x^{2} +3x)}{x(x-3)(x+3)} =0\\\\\\\frac{6x^{2} +14x+7x-21-6x^{2} -18x}{x(x-3)(x+3)} =0\\\\\\\frac{3x-21}{x(x-3)(x+3)} =0\\\\\\\frac{3\cdot(x-7)}{x(x-3)(x+3)}=0\\\\\\\left \{ {{x-7=0} \atop {x\neq 0 \ ; \ x-3\neq 0 \ ; \ x+3\neq 0}} \right.$

$\displaystyle\bf\\\left \{ {{x=7} \atop {x\neq 0 \ ; \ x\neq 3 \ ; \ x\neq -3}} \right. \\\\\\Otvet \ : \ 7$