Question

Решите 16 пожалуйста

Answer 1

$\frac{1}{x^2+2x-3} +\frac{18}{x^2+2x+2} =\frac{18}{x^2+2x+1} \\\\ODZ:\; \left\{\begin{array}{c}x^2+2x-3\ne 0\\x^2+2x+2\ne 0\\x^2+2x+1\ne 0\end{array}\right \; ,\; \left\{\begin{array}{c}(x-1)(x+3)\ne 0\\x^2+2x+2\ \textgreater \ 0\; pri\; x\in R\\(x+1)^2\ne 0\end{array}\right \\\\x\ne 1\; ,\; x\ne -3\; ,\; x\ne -1\\\\Zamena:\; \; t=x^2+2x\\\\\frac{1}{t-3} +\frac{18}{t+2} = \frac{18}{t+1} \\\\ \frac{(t+2)(t+1)+18(t-3)(t+1)-18(t-3)(t+2)}{(t-3)(t+2)(t+1)} =0$

$t^2+3t+2+18(t^2-2t-3)-18(t^2-t-6)=0$

$t^2-15t+56=0\\\\D=225-224=1\\\\t_1=8\; ,\; \; t_2=7\\\\a)\; \; x^2+2x=8\\\\x^2+2x-8=0\\\\D/4=9\; ,\; x_1=-4\; ,\; x_2=2\\\\b)\; \; x^2+2x=7\\\\x^2+2x-7=0\\\\D/4=8\\\\x_3= -1-\sqrt8\; ,\; \; x_4=-1+\sqrt8$