Question

. 50б решить. Уравнение высших степеней.

Answer 1

Объяснение:

$\displaystyle\bf\\x^2+\Bigg(\frac{x}{2x-1}\Bigg)^2=2;\ \ \ \ \ \ \ \ x\neq \frac{1}{2}\\\\\\x^2+\frac{x^2}{(2x-1)^2} =2\\\\x^2\Bigg(1+\frac{1}{(2x-1)^2} \Bigg)=2\\\\\\\frac{4x^2-4x+1+1}{(2x-1)^2} =\frac{2}{x^2} \\\\\\\frac{2x^2-2x+1}{(2x-1)^2}=\frac{1}{x^2} \\\\\\\frac{x^2-2x+1}{(2x-1)^2}+\frac{x^2}{(2x-1)^2} =\frac{1}{x^2} \\\\\\ \frac{(x-1)^2}{(2x-1)^2}=\frac{1}{x^2} -\frac{x^2}{(2x-1)^2} \\\\\\\frac{(x-1)^2}{(2x-1)^2}=\frac{(2x-1)^2-(x^2)^2}{x^2(2x-1)^2}$

$\displaystyle\bf\\\frac{(x-1)^2}{(2x-1)^2}=\frac{(2x-1-x^2)(2x-1+x^2)}{x^2(2x-1)^2} \\\\\\\frac{(x-1)^2}{(2x-1)^2}=-\frac{(x-1)^2(x^2+2x-1)}{x^2(2x-1)^2} \\\\\\(x-1)^2\cdot\Bigg(\frac{1}{(2x-1)^2}+\frac{x^2+2x-1}{x^2(2x-1)^2} \Bigg)=0\\\\\\(x-1)^2\cdot\frac{x^2+x^2+2x-1)}{x^2(2x-1)^2} =0\\\\\\(x-1)^2\cdot( 2x^2+2x-1)=0\\\\1)\ (x-1)^2=0;\ \ x_1=1\\\\2)\ 2x^2+2x-1=0\\\\D=4+8=12\\\\x_{2,3}=\frac{-2\pm2\sqrt{3} }{4} =\frac{-1\pm\sqrt{3} }{2} \\\\\\Otvet:1;\ \frac{-1\pm\sqrt{3} }{2}$