Question

Решите уравнение 4(x^2+1) / (x^2+10x+1) - 5x/(x^2+1) + 3.5 = 0

Answer 1

$\frac{4(x^2+1)}{x^2+10+1}- \frac{5x}{x^2+1} + \frac{7}{2}=0\\\\ \frac{ 4(\frac{x^2+1}{x^2+1}) }{ \frac{x^2+1+10x}{x^2+1} }- \frac{5x}{x^2+1}+ \frac{7}{2}=0\\\\ \frac{4}{ 1+\frac{10x}{x^2+1} } - \frac{5x}{x^2+1}+ \frac{7}{2}=0\\\\ \frac{5x}{x^2+1}=t\\\\ \frac{4}{1+2t}- \frac{t}{1}+ \frac{7}{2}=0\\\\ 8-2t-4t^2+7+14t=0\\ -4t^2+12t+15=0\\ 4t^2-12t-15=0\\ D=144+240=384;\ \sqrt{D}= \sqrt{384}= \sqrt{16 \cdot 4\cdot 6}=4 * \sqrt{6}=8 \sqrt{6}\\\\ t_{1/2}= \frac{12\pm 8 \sqrt{6} }{8}=\frac{3\pm 2 \sqrt{6} }{2} = \frac{3}{2}+\sqrt {6}$

Обратная замена

$\frac{5x}{x^2+1}= \frac{3}{2} \pm \sqrt6\\\\ \frac{5x}{x^2+1}= \frac{3 \pm 2\sqrt 6}{2}\\\\ (x^2+1)(3\pm2sqrt6)=10x\\ 3\pm2\sqrt6)x^2-10x+(3\pm2\sqrt6)=0\\ D=100-4(3\pm2\sqrt6)^2=100-4(9\pm12\sqrt6+24)=\\ =100-36\mp48\sqrt6-96=-32\mp48\sqrt6\\ D_1=-21-48\sqrt6\ \textless \ 0; D_2=48\sqrt6-32\ \textgreater \ 0\\\\ x_{1/2}= \frac{10\pm \sqrt{48\sqrt6-32}}{2(3-2\sqrt6)}= \frac{10\pm 4 \sqrt{3\sqrt6-2}}{2 (3-2\sqrt6)}= \frac{5\pm2\sqrt{3\sqrt6-2}}{3-2\sqrt6}$