Question

1 supra x + x+1 supra x-3= 12 supra x(x-3)

Answer 1

$\frac{1}{x}+\frac{x+1}{x-3}=\frac{12}{x(x-3)}\\\\\text{C.E.} \left \{ {{x \neq 0} \atop {x-3 \neq 0}} \right. \\\\\frac{x-3}{x(x-3)}+\frac{x(x+1)}{x(x-3)}=\frac{12}{x(x-3)}\\\\x-3+x^2+x=12\\x^2+2x-15=0\\x^2+5x-3x-15=0\leftrightarrow x(x+5)-3(x+5)=0\leftrightarrow(x+5)(x-3)=0\\\leftrightarrow(x=-5\text{ sau }x=3\text{ nu verific\u a C.E.})\\x\in\{-5\}$