Question

решите уравнение......​

Answer 1

$\displaystyle \frac{2x-5}{x^2-3x}+\frac{x-5}{x^2-9}=\frac{x+2}{x^2+3x},x\neq 0,x\neq 3,x\neq -3\\ \\\frac{2x-5}{x^2-3x}+\frac{x-5}{x^2-9}-\frac{x+2}{x^2+3x}=0\\ \\\frac{2x-5}{x(x-3)}+\frac{x-5}{(x-3)(x+3)}-\frac{x+2}{x(x+3)}=0\\ \\\frac{(x+3)(2x-5)+x(x-5)-(x-3)(x+2)}{x(x-3)(x+3)}=0\\ \\\frac{2x^2-5x+6x-15+x^2-5x-(x^2-x-6)}{x(x-3)(x+3)}=0\\ \\\frac{2x^2-3x-9}{x(x-3)(x+3)}=0\\ \\\frac{2x+3}{x(x+3)}=0\\ \\2x+3=0\\ 2x=-3\\ \\x=-\frac{3}{2}=-1\frac{1}{2}=-1,5$