Question

55 , нужно развёрнутое решение

Answer 1

$\left \{ {{2xy-1=\frac{y^3}{x}} \atop {3xy-2=\frac{x^3}{y}}} \right. \; \; \; \; \; \; ODZ:\; x\ne 0\; ,\; y\ne 0\\\\\\\left \{ {{2x^2y-x=y^3} \atop {3xy^2-2y=x^3}} \right.\; \; \to \; \; (2x^2y-x)(3xy^2-2y)=x^3y^3\\\\6x^3y^3-4x^2y^2-3x^2y^2+2xy-x^3y^3=0\\\\5(xy)^3-7(xy)^2+2(xy)=0\\\\xy\cdot \Big (5(xy)^2-7(xy)+2\Big )=0\; \; ,\; \; xy\ne 0\\\\t=xy\; \; ,\; \; 5t^2-7t+2=0\; \; ,\; \; t_1=1\; ,\; t_2=\frac{2}{5}\\\\1)\; \; xy=1\; \; ,\; \; x=\frac{1}{y}\\\\2xy-1=2\cdot 1-1=1\; \; \to \; \; 1=\frac{y^3}{x}\; \; ,\; \; x=y^3\; \; ,\; \; y^3-\frac{1}{y}=0\\\\\frac{y^4-1}{y}=0\; \; ,\; \; \frac{(y-1)(y+1)(y^2+1)}{y}=0\; \; ,\; \; y_1=1\; ,\; y_2=-1\\\\x_1=1\; ,\; \; x_2=-1$

$2)\; \; xy=\frac{2}{5}\; \; ,\; \; x=\frac{2}{5y}\\\\2xy-1=2\cdot \frac{2}{5}-1=-\frac{1}{5}\; \; ,\; \; -\frac{1}{5}=\frac{y^3}{x}\; \; ,\; \; -\frac{1}{5}=\frac{5y\cdot y^3}{2}\; \; ,\; \; \frac{5y^4}{2}+\frac{1}{5}=0\; ,\\\\\frac{25y^4+2}{10}=0\; \; \to \; \; \; 25y^4+2=0\; \; \to \; \; y\in \varnothing ,\; tak\; kak\; \; \; 25y^4+20\\\\Otvet:\; \; (1,1)\; ,\; (-1,-1)\; .$