Question

Помогите пж!!! Никак не могу понять.​

Answer 1

$\displaystyle\bf\\\frac{6x-3a-a^{2} }{ax} =1+a\\\\\\6x-3a-a^{2} =ax\cdot(1+a)\\\\\\6x-3a-a^{2} =ax+a^{2} x\\\\\\a^{2} x+ax-6x=-a^{2} -3a\\\\\\x\cdot(a^{2} +a-6)=-a^{2} -3a\\\\\\x\cdot(a+3)\cdot(a-2)=-a\cdot(a+3)\\\\\\x=\frac{-a\cdot(a+3)}{(a+3)\cdot(a-2)} =\frac{-a}{a-2} =\frac{a}{2-a} \\\\\\Otvet \ : \ x=\frac{a}{2-a}$