Question

помогите пожалуйста!! надо решить систему!

Answer 1

Объяснение:

Пусть  $\frac{1}{x^2+3xy} =t,\ \ \ \ \frac{1}{y^2-xy}=v\ \ \ \ \ \Rightarrow$

$\left \{ {{2t+3v=\frac{25}{14}\ |*14 } \atop {3t-2v=-\frac{4}{7}\ |*21 }} \right.\ \ \ \ \left \{ {{28t+42v=25} \atop {63t-42v=-12}} \right. .$

Суммируем эти уравнения:

$91t=13\ |:91\\t=\frac{1}{x^2+3xy} =\frac{1}{7} .\\3*\frac{1}{7} -2v=-\frac{4}{7} \\2v=1\ |:2\\v=\frac{1}{y^2-xy}= \frac{1}{2}. \ \ \ \ \Rightarrow\\\left \{ {{\frac{1}{x^2+3xy} =\frac{1}{7} } \atop {\frac{1}{y^2-xy}=\frac{1}{2} }} \right.\ \ \ \ \left \{ {{x^2+3xy=7} \atop {y^2-xy=2}} \right. .$

Суммируем эти уравнения:

$x^2+2xy+y^2=9\\(x+y)^2=9\\1.\\\left \{ {{x+y=3} \atop {y^2-xy=2}} \right.\ \ \ \ \left \{ {{x=3-y} \atop {y^2-(3-y)*y=2}} \right. \ \ \ \ \ \left \{ {{x=3-y} \atop {y^2-3y+y^2=2}} \right.\ \ \ \ \left \{ {{x=3-y} \atop {2y^2-3y-2=0}} \right. \\\left \{ {{x=3-y} \atop {D=25\ \ \ \ \sqrt{D}=5 }} \right. \ \ \ \ \left \{ {{x_1=1\ \ \ \ x_2=3,5} \atop {y_1=2\ \ \ \ y_2=-0,5}} \right..$

$2.\\\left \{ {{x+y=-3} \atop {y^2-xy=2}} \right. \ \ \ \ \left \{ {{x=-3-y} \atop {y^2-(-3-y)*y=2}} \right.\ \ \ \ \left \{ {{x=-3-y} \atop {y^2+3y+y^2=2}} \right.\ \ \ \ \left \{ {{x=-3-y} \atop {2y^2+3y-2=0\\}} \right. \\\left \{ {{x=-3-y} \atop {D=25\ \ \ \ \sqrt{D}=5 }} \right. \ \ \ \ \left \{ {{x_3=-1\ \ \ \ x_4=-3,5} \atop {y_3=-2\ \ \ \ y_4=0,5}} \right. .$

Ответ: (1;2),  (3,5;-0,5),  (-1;-2),  (-3,5;0,5).

Answer 2

$\left\{\begin{array}{ccc}\dfrac{2}{x^{2}+3xy }+\dfrac{3}{y^{2}-x y}=\dfrac{25}{14} \\\dfrac{3}{x^{2}+3xy }-\dfrac{2}{y^{2}-x y}=-\dfrac{4}{7} \end{array}\right\\\\\\\dfrac{1}{x^{2}+3xy }=m \ ; \ \dfrac{1}{y^{2}-x y}=n\\\\\\\left\{\begin{array}{ccc}2m+3n=\dfrac{25}{14} \ |\cdot2 \\3m-2n=-\dfrac{4}{7} \ |\cdot3 \end{array}\right \\\\\\+\left\{\begin{array}{ccc}4m+6n=\dfrac{25}{7}\\9m-6n=-\dfrac{12}{7} \end{array}\right\\-----------\\13m=\dfrac{13}{7} \\\\m=\dfrac{1}{7}$

$2n=3m+\dfrac{4}{7}=3\cdot\dfrac{1}{7}+\dfrac{4}{7} =\dfrac{3}{7}+\dfrac{4}{7}=1\\\\n=\dfrac{1}{2}\\\\\\\left\{\begin{array}{ccc}\dfrac{1}{x^{2}+3xy }=\dfrac{1}{7} \\\dfrac{1}{y^{2} -xy}=\dfrac{1}{2}\end{array}\right\\\\\\+\left\{\begin{array}{ccc}x^{2}+3xy=7 \\y^{2}-xy=2 \end{array}\right \\---------\\x^{2} +2xy+y^{2}=9\\\\(x+y)^{2} =9\\\\x+y=\pm3\\\\1)\\\left\{\begin{array}{ccc}x+y=-3\\y^{2}-xy=2 \end{array}\right$

$\left\{\begin{array}{ccc}x=-y-3\\y^{2}-(-y-3)\cdot y=2 \end{array}\right\\\\\left\{\begin{array}{ccc}x=-y-3\\y^{2}+y^{2} +3y-2=0 \end{array}\right\\\\\left\{\begin{array}{ccc}x=-y-3\\2y^{2}+3y-2=0 \end{array}\right\\\\\left\{\begin{array}{ccc}x=-y-3\\\left[\begin{array}{ccc}y_{1} =-2\\y_{2}=0,5 \end{array}\right \end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{1} =-1} \atop {y_{1}=-2 }} \right. \\\left \{ {{x_{2}=-3,5 } \atop {y_{2} =0,5}} \right. \end{array}\right$

$2)\\\left\{\begin{array}{ccc}x+y=3\\y^{2}-xy=2 \end{array}\right\\\\\\\left\{\begin{array}{ccc}x=3-y\\y^{2}-(3-y)\cdot y=2 \end{array}\right\\\\\left\{\begin{array}{ccc}x=3-y\\y^{2}-3y+y^{2}-2=0 \end{array}\right\\\\\left\{\begin{array}{ccc}x=3-y\\2y^{2}-3y-2=0 \end{array}\right$

$\left\{\begin{array}{ccc}x=3-y\\\left[\begin{array}{ccc}y_{3} =2\\y_{4}=-0,5 \end{array}\right \end{array}\right\\\\\\\left[\begin{array}{ccc}\left \{ {{x_{3} =1} \atop {y_{3}=2 }} \right. \\\left \{ {{x_{4}=3,5 } \atop {y_{4} =-0,5}} \right. \end{array}\right\\\\\\Otvet:\boxed{(-1 \ ; \ -2) \ ; \ (-3,5 \ ; \ 0,5) \ ; \ (1 \ ; \ 2) \ , \ (3,5 \ ; \ -0,5)}$