Giải hệ pt: a) 3y= (y ² +2)/x ² 3x= (x ² +2)/y ² b) x- (1/x)=y- (1/y) xy= x ³ +1

a) Lay ptrinh(1) – ptrinh (2) ta co

$3(y-x) = \dfrac{(y^2-x^2)(x^2 + y^2 + 2)}{x^2 y^2}$

<->$ 3 (y-x) = \dfrac{(y-x)(y+x)(x^2 + y^2 + 2)}{x^2 y^2}$

<->$y=x$ hoac $3 = \dfrac{(y+x) (x^2 + y^2 + 2)}{x^2 y^2}$.

TH1: y = x

Thay vao ptrinh (1) ta co

$3x = (x^2+2)/x^2$ hay $(x-1)(3x^2 + 2x + 2) = 0$

Vay $x=1$, suy ra $y= 1$. Vay nghiem la $(1,1)$.

TH2: $3 = \dfrac{(y+x)(x^2 + y^2 + 2)}{x^2 y^2}$

b)Ptrinh dau tien tuong duong vs

$x-y+\dfrac{1}{y} – \dfrac{1}{x} = 0$

$(x-y) + \dfrac{x-y}{xy} = 0$

$(x-y)(1+\dfrac{1}{xy}) = 0$

Vay $x=y$ hoac $xy=-1$

TH1: x=y

THay vao ptrinh sau ta co

$x^3-x^2+1=0$

TH2: xy = -1 hay y = -1/x

THay vao ptrinh sau ta co

$x.(-1/x) = x^3+1$

<->$ -1 = x^3+1$

<-> $x^3 = -2$

<-> $x = \sqrt[3]{-2}$

Vay $y = \dfrac{\sqrt[3]{4}}{-2}$

Vay nghiem la $(\sqrt[3]{-2}, \dfrac{\sqrt[3]{4}}{-2})$.