giai he pt y^2 + 2xy +4 =2x+5y 5x^2+7y-18= √x^4 +4 19/07/2021 Bởi Eden giai he pt y^2 + 2xy +4 =2x+5y 5x^2+7y-18= √x^4 +4
$\left \{ {{y^2 + 2xy +4 =2x+5y} \atop {5x^2+7y-18= \sqrt{x^4+4}}} \right.$ ⇔$(y^2-y)+(4-4y)+2x(y-1)=0$ ⇔$(y-1)(y-4+2x)=0$ ⇔\(\left[ \begin{array}{l}y=1\\y=4-2x\end{array} \right.\) TH1: y=1 pt (2) trở thành: $5x^2+7-18=\sqrt{5}⇔x^2=\frac{11+\sqrt{5}}{5}$⇒x=… TH2: y=4-2x pt(2) trở thành: $5x^2+7(4-2x)-18=\sqrt{x^4+4}$ ⇔$5x^2-14x+10=\sqrt{x^4+4}$ ⇔$25x^4+196x^2+100-140x^3+100x^2-280x=x^4+4$ ⇔$24x^4-140x^3+296x^2-280x+96$=0 Bình luận
$\left \{ {{y^2 + 2xy +4 =2x+5y} \atop {5x^2+7y-18= \sqrt{x^4+4}}} \right.$
⇔$(y^2-y)+(4-4y)+2x(y-1)=0$
⇔$(y-1)(y-4+2x)=0$
⇔\(\left[ \begin{array}{l}y=1\\y=4-2x\end{array} \right.\)
TH1: y=1
pt (2) trở thành: $5x^2+7-18=\sqrt{5}⇔x^2=\frac{11+\sqrt{5}}{5}$⇒x=…
TH2: y=4-2x
pt(2) trở thành: $5x^2+7(4-2x)-18=\sqrt{x^4+4}$
⇔$5x^2-14x+10=\sqrt{x^4+4}$
⇔$25x^4+196x^2+100-140x^3+100x^2-280x=x^4+4$
⇔$24x^4-140x^3+296x^2-280x+96$=0