Question

решите прошу Ну очень надо

Answer 1

$1) x+3+ \frac{18}{x-3}= \frac{ x^{2} -9+18}{x-3} = \frac{ x^{2} +9}{x-3}\\\\\\2) \frac{ x^{2} +9}{x-3} * \frac{2 x^{2} -12x+18}{ x^{2} +9}= \frac{2( x^{2} -6x+9)}{x-3}= \frac{2(x-3) ^{2} }{x-3} =2x-6$

$\frac{(x-3) ^{2} }{12}*12- \frac{2 x^{2} +5}{4}*12+ \frac{1,25 x^{2} -1}{3}*12= \frac{2}{3}*12\\\\ x^{2} -6x+9-6 x^{2} -15+5 x^{2} -4=8\\\\-6x=18\\\\x=-3$