Question

пж! срочно нужно ответ

​

Answer 1

$\displaystyle \frac{12-x}{x^2+6x}+\frac{3}{x^2-6x}=\frac{6}{x^2-36},x\neq 0,x\neq -6,x\neq 6\\ \\\frac{12-x}{x^2+6x}+\frac{3}{x^2-6x}-\frac{6}{x^2-36}=0\\ \\\frac{(x-6)(12-x)+3(x+6)-6x}{x(x-6)(x+6)}=0\\ \\\frac{12x-x^2-72+6x+3x+18-6x}{x(x-6)(x+6)}=0\\ \\\frac{-x^2+9x+6x-54}{x(x-6)(x+6)}=0\\ \\\frac{-x(x-9)+6(x-9)}{x(x-6)(x+6)}=0\\ \\\frac{-(x-9)}{x(x+6)}=0\\ \\-\frac{x-9}{x(x+6)}=0\\ \\\frac{9-x}{x(x+6)}=0\\ \\9-x=0\\ -x=-9 \\ x=9$