Question

Даю 40 баллов! Срочно!

Answer 1

$\displaystyle\bf\\\frac{a-c-5}{x} =\frac{a^{2} -10a+25-c^{2} }{a^{2} +2ac+c^{2} -5a-5c} \\\\\\\frac{a-c-5}{x} =\frac{(a^{2} -10a+25)-c^{2} }{(a^{2} +2ac+c^{2} )-5\cdot(a+c)} \\\\\\\frac{a-c-5}{x} =\frac{(a-5)^{2} -c^{2} }{(a+c)^{2} -5\cdot(a+c)} \\\\\\\frac{a-c-5}{x} =\frac{(a-5-c)\cdot(a-5+c)}{(a+c)\cdot(a+c-5)} \\\\\\\frac{a-c-5}{x} =\frac{a-c-5}{a+c} \\\\\\x=\frac{(a-c-5)\cdot(a+c)}{a-c-5} =a+c\\\\\\Otvet \ : \ x=a+c$