\begin{cases}y(x^{2}+1)=2x(y^{2}+1) \\ (x^{2}+y^{2})(1+\frac{1}{x^{2}y^{2}})=24 \end{cases}
pt: $(x^2+y^2)(1+1/x^2.y^2)=24$      (2)
<=>$x^2+1+y^2+1+1/x^2+1+1/y^2+1=20$
<=>$x^2+1+(x^2+1)/x^2+y^2+1+(y^2+1)y^2=20$
<=>$(x^2+1)(1+1/x^2)+(y^2+1)(1+1/y^2)=20$
<=>$(x^2+1)[(x^2+1)/x^2]+(y^2+1)[(y^2+1)/y^2]=20$
<=>$[(x^2+1)^2]/x^2+(y^2+1)^2/y^2=20$
<=>$[y(x^2+1)]^2+[x(y^2+1)]^2=20x^2.y^2$  (quy đồng khử mẫu đó mà)    (3)
pt: $y(x^2+1)=2x(y^+1)$     (1)
thay (1) vào (3) ta được: $[2x(y^+1)]^2+[x(y^2+1)]^2=20x^2.y^2$
<=> $5[x(y^2+1)]^2=20x^2.y^2$
<=> $[x(y^2+1)]^2=4x^2.y^2$
<=> $[x(y^2+1)]^2 - 4x^2.y^2=0$
<=>$[x(y^2+1) - 2xy][x(y^2+1) + 2xy]=0$
<=>$x^2(y^2+1 - 2y)(y^2+1+2y)=0$
đến đây tìm y sau đó quay lại tìm x bạn nhé!
khó nhìn quá bạn ơi :( –  dolaemon 23-08-15 07:32 PM

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