Question

решите уравнение, пожалуйста ​

Answer 1

$\displaystyle \frac{4}{x^2-6x+9}-\frac{6}{x^2-9}=\frac{1}{x+3},x\neq 3,x\neq-3 \\\\\frac{4}{x^2-6x+9}-\frac{6}{x^2-9}-\frac{1}{x+3}=0\\ \\\frac{4}{(x-3)^2}-\frac{6}{(x-3)(x+3)}-\frac{1}{x+3}=0\\ \\\frac{4(x+3)-6(x-3)-(x-3)^2}{(x-3)^2(x+3)}=0\\ \\\frac{4x+21-x^2}{(x-3)^2(x+3)}=0\\ \\\frac{-x(x-7)-3(x-7)}{(x-3)^2(x+3)}=0\\ \\\frac{-x(x-7)}{(x-3)^2}=0\\ \\-\frac{x-7}{(x-3)^2}=0\\ \\\frac{7-x}{(x-3)^2}=0\\ \\7-x=0\\-x=-7\\x=7$