Toán 9 Giúp tớ câu hệ

D

dien0709

Giải hệ phương trình
pt(1) : $x^3 -y^3 =4x+2y$
pt(2) :$ x^2- 1 = 3(1-y^2)$

(2)=>$4=x^2+3y^2$(3) <=> $x^2-4=-3y^2=>|x|\le 2$

(1)=>$x^3-y^3=x(x^2+3y^2)+2y$

=>$y=0=>x=\pm 2$

$y^2+3xy+2=0=>x=\dfrac{-y^2-2}{3y}$

$(3)=>7y^4-8y^2+1=0=>y=\pm 1 ; y=\pm \dfrac{\sqrt{7}}{7}$

KL:$(x;y)=(2;0) , (-2;0) , (1;-1) , (-1;1)$

=$ ( \dfrac{\sqrt{7}}{7};-\dfrac{5\sqrt{7}}{7}) , (-\dfrac{\sqrt{7}}{7};\dfrac{5\sqrt{7}}{7})$
 
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