{x^2y^2-xy=12- {x+y=2

24 ноября 2013 / Алгебра

{x^2y^2-xy=12;  {x+y=2.

  • x=2-y\end{cases}\\\\xy(xy-1)-12=0\\(2-y)*y*((2-y)*y-1)-12=0\\((2-y)*y)^2-(2-y)*y-12=0\\(2-y)*y=a\\a^2-a-12=0\\a_1=4\ ;a_2=-3\\(2-y)*y=4\ ;\ \ \ \ \ \ \ \ \ \ (2-y)*y=-3\\y^2-2y+4=0\ ;\ \ \ \ \ \ \ \ \ y^2-2y-3=0\\y_{1,2}=1^+_-\sqrt{1-4}\ ;\ \ \ \ \ \ \ \ y_{3,4}=1^+_-2\\error\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y_3=3\ y_4=-1\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_3=-1\ x_4=3′ alt=’\begin{cases}x^2*y^2-xy=12\\x+y=2=>x=2-y\end{cases}\\\\xy(xy-1)-12=0\\(2-y)*y*((2-y)*y-1)-12=0\\((2-y)*y)^2-(2-y)*y-12=0\\(2-y)*y=a\\a^2-a-12=0\\a_1=4\ ;a_2=-3\\(2-y)*y=4\ ;\ \ \ \ \ \ \ \ \ \ (2-y)*y=-3\\y^2-2y+4=0\ ;\ \ \ \ \ \ \ \ \ y^2-2y-3=0\\y_{1,2}=1^+_-\sqrt{1-4}\ ;\ \ \ \ \ \ \ \ y_{3,4}=1^+_-2\\error\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ y_3=3\ y_4=-1\\.\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_3=-1\ x_4=3′ align=’absmiddle’ class=’latex-formula’>


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