Question

100 баллов. Решите систему уравнений... С решением пж​

Answer 1

$\displaystyle\left \{ {{x^{4}+y^{4} =17 } \atop {x+y=3}} \right. \\\\x^{4} +y^{4}=17\\\\(x^{2} +y^{2} )^{2} -2x^{2} y^{2}=17\\\\\Big(\underbrace{(x+y}_{3})^{2} -2xy\Big)^{2} -2x^{2}y^{2} =17\\\\(9-2xy)^{2} -2(xy)^{2} =17\\\\xy=m\\\\(9-2m)^{2} -2m^{2} -17=0\\\\81-36m+4m^{2}-2m^{2} -17=0\\\\2m^{2} -36m+64=0\\\\m^{2} -18m+32=0$

$m_{1} =2 \ ; \ m_{2} =16\\\\1)\\\displaystyle\left \{ {{x+y=3} \atop {xy=2}} \right.\\\\\\\left[\begin{array}{ccc}\left \{ {{x=1} \atop {y=2}} \right. \\\\\left \{ {{x=2} \atop {y=1}} \right. \end{array}\right\\\\\\2)\\\displaystyle\left \{ {{x+y=3} \atop {xy=16}} \right. \\\\\\\displaystyle\left \{ {{y=3-x} \atop {x(3-x)-16=0}} \right.\\\\\\\displaystyle\left \{ {{y=3-x} \atop {x^{2} -3x+16=0}} \right.\\\\\\x^{2} -3x+16=0\\\\D=(-3)^{2} -4\cdot16=9-64=-55<0\\\\x\in\oslash$

$Otvet:\boxed{(1 \ ; \ 2) \ , \ (2 \ ; \ 1)}$