Question

помогите решить этот пример даю 40 баллов

Answer 1

$\displaystyle\bf\\\frac{x^{2} +(3-a)x-3a}{x^{2} -x-12} =0\\\\\\\left \{ {{x^{2} +(3-a)x-3a=0} \atop {x^{2} -x-12\neq 0}} \right. \\\\\\\left \{ {{x^{2} +(3-a)x-3a=0} \atop {(x-4)(x+3)\neq 0}} \right.\\\\\\\left \{ {{x^{2} +(3-a)x-3a=0} \atop {x\neq 4 \ ; \ x\neq -3}} \right.\\\\\\x^{2} +(3-a)x-3a=0\\\\\\D=(3-a)^{2} -4\cdot (-3a)=9-6a+a^{2} +12a=a^{2}+6a+9=(a+3)^{2}\\\\\\x_{1} =\frac{-(3-a)+(a+3)}{2} =\frac{-3+a+a+3}{2}=a$

$\displaystyle\bf\\x_{2} =\frac{-(3-a)-(a+3)}{2} =\frac{-3+a-a-3}{2} =\frac{-6}{2} =-3-neyd\\\\\\Otvet:x=a$