Question

решите дробно-рациональное уравнение​

Answer 1

$\displaystyle \frac{x-5}{x+3}-\frac{8}{x-3}=\frac{48}{x^2-9},x\neq -3,x\neq 3\\\\\frac{x-5}{x+3}-\frac{8}{x-3}-\frac{48}{x^2-9}=0\\\\\frac{x-5}{x+3}-\frac{8}{x-3}-\frac{48}{(x-3)(x+3)}=0\\\\\frac{(x-3)(x-5)-8(x+3)-48}{(x-3)(x+3)}=0\\\\\frac{x^2-5x-3x+15-8x-24-48}{(x-3)(x+3)}=0\\\\\frac{x^2-16x-57}{(x-3)(x+3)}=0\\\\\frac{x^2+3x-19x-57}{(x-3)(x+3)}=0\\\\\frac{x(x+3)-19(x+3)}{(x-3)(x+3)}=0\\\\\frac{(x+3)(x-19)}{(x-3)(x+3)}=0\\\\\frac{x-19}{x-3}=0\\\\x-19=0\\x=19$