Question

решить уравнение методом замены переменной
$2(\frac{2+x}{x-1})^{2} - 7(\frac{2+x}{x-1})+5=0$

Answer 1

$\displaystyle\tt 2\bigg(\frac{2+x}{x-1}\bigg)^2-7\bigg(\frac{2+x}{x-1}\bigg)+5=0\\\\\\ \bold{\frac{2+x}{x-1}=t}\\\\ 2t^2-7t+5=0\\D=49-40=9=3^2\\t_1=\frac{7-3}{4}=1\\t_2=\frac{7+3}{4}=2.5\\\\\\ 1) \ \frac{2+x}{x-1}=1; \ \ \ x\neq 1\\ 2+x=x-1\\2\neq -1\\x\in\O\\\\\\ 2) \ \frac{2+x}{x-1}=2.5; \ \ \ x\neq 1\\ 2+x=2.5(x-1)\\2+x=2.5x-2.5\\1.5x=4.5\\x=3$

Ответ: х = 3.