Question

Решите уравнение !
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Answer 1

$\frac x{x+3}-\frac{18}{x^2-9}=\frac{x+3}{3-x}\\\\O.D.3.:\\\begin{cases}x\neq-3\\x\neq3\end{cases}\\\\\frac x{x+3}-\frac{18}{x^2-9}-\frac{x+3}{3-x}=0\\\\\frac x{x+3}-\frac{18}{(x-3)(x+3)}+\frac{x+3}{x-3}=0\\\\\frac{x(x-3)-18+(x+3)(x+3)}{(x-3)(x+3)}=0\\\\\frac{x^2-3x-18+x^2+6x+9}{x^2-9}=0\\\\\frac{2x^2+3x-9}{x^2-9}=0\\\\2x^2+3x-9=0\\D=9-4\cdot2\cdot(-9)=9+72=81\\x_{1,2}=\frac{-3\pm9}4\\x_1=-3\;-\;He\;nogx.\;no\;O.D.3.\\x_2=1,5$