Question

Решите систему уравнений (во вложении)

Answer 1

ОДЗ:  $\boxed{x\ \textgreater \ 0}$

$\frac{x^2}{4} -xy \geq 0; \ \ \ x^2 \geq 4xy; \ \ x \geq 4y \ \Rightarrow \ x\ \textgreater \ 0, \ y \leq \frac{x}{4} \\ \\ x+4y^2 \geq 0; \ \ \ x \geq -4y^2 \ \Rightarrow \ x\ \textgreater \ 0 \\ \\ x-2y \geq 0; \ \ x \geq 2y \ \Rightarrow \ x\ \textgreater \ 0, \ y \leq \frac{x}{2}$

$x+4y^2 =(x-2y)^2 \\ \\ x+4y^2 =x^2 -4xy+4y^2 \\ \\ 4xy =x^2 -x; \ \ \ y = \frac{1}{4x} \cdot x(x -1) = \frac{x-1}{4}$


$\log_2^2 x + \log_2 (\frac{x^2}{4} - x \cdot \frac{x-1}{4})=0 \\ \\ \log_2^2 x + \log_2 (\frac{x^2}{4} -\frac{x^2-x}{4})=0; \ \ \ \ \ \log_2^2 x + \log_2 (\frac{x^2}{4} -(\frac{x^2}{4}- \frac{x}{4}))=0 \\ \\ \log_2^2 x + \log_2 \frac{x}{4}=0; \ \ \ \log_2^2 x + \log_2 x - \log_24=0 \\ \\ t= \log_2 x \\ \\ t^2 +t-2=0; \ \ \ t_{1,2}=\frac{-1 \pm \sqrt{1+8}}{2} = \frac{-1 \pm 3}{2}; \ \ \ t_1=1; \ t_2=-2 \\ \\ \log_2x=1; \ \ x=2^1=2; \\ \\ \log_2x=-2; \ \ x=2^{-2}=\frac{1}{4}$

$y=\frac{2-1}{4}=\frac{1}{4} \ \Rightarrow \ (2; \frac{1}{4}) \\ \\ y=\frac{\frac{1}{4}-1}{4}=-\frac{3}{16} \ \Rightarrow \ (\frac{1}{4}; -\frac{3}{16})$