Đáp án:
`(x;y)\in {(0;0);(1;1/ 2);(-1;-1/ 2);(\sqrt{5/ 2};\sqrt{10});(-\sqrt{5/ 2};-\sqrt{10})}`
Giải thích các bước giải:
$\quad \begin{cases}2x^2-5xy+2y^2=0\ (1)\\x^2y=3y-x\ (2)\end{cases}$
`(1)<=>2x^2-4xy-xy+2y^2=0`
`<=>2x(x-2y)-y(x-2y)=0`
`<=>(x-2y)(2x-y)=0`
`<=>`$\left[\begin{array}{l}x-2y=0\\2x-y=0\end{array}\right.$
$\\$
+) Với `x-2y=0<=>x=2y`
`(2)<=>(2y)^2 y =3y-2y`
`<=>4y^3-y=0`
`<=>y(4y^2-1)=0`
`<=>`$\left[\begin{array}{l}y=0\\4y^2=1\end{array}\right.$
`<=>`$\left[\begin{array}{l}y=0\\y=\dfrac{1}{2}\\y=\dfrac{-1}{2}\end{array}\right.$`=>`$\left[\begin{array}{l}x=0\\x=2y=2.\dfrac{1}{2}=1\\x=2y=2.(\dfrac{-1}{2})=-1\end{array}\right.$
$\\$
+) Với `2x-y=0<=>2x=y`
`(2)<=>x^2 .2x=3.2x-x`
`<=>2x^3-5x=0`
`<=>x(2x^2-5)=0`
`<=>`$\left[\begin{array}{l}x=0\\2x^2=5\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=0\\x=\sqrt{\dfrac{5}{2}}\\x=-\sqrt{\dfrac{5}{2}}\end{array}\right.$`=>`$\left[\begin{array}{l}y=0\\y=2x=\sqrt{10}\\y=-\sqrt{10}\end{array}\right.$
Vậy hệ phương trình có nghiệm:
`(x;y)\in {(0;0);(1;1/ 2);(-1;-1/ 2);(\sqrt{5/ 2};\sqrt{10});(-\sqrt{5/ 2};-\sqrt{10})}`
