( toán 10)- hệ phương trình...

H

hoangtrongminhduc

(1)<=>$y^2=\dfrac{2x}{x^2+1} $\Rightarrow $\left\{\begin{matrix}y \in [-1;1] \\ x \ge0\end{matrix}\right.$(*)
thay (*) vào (2) ta có:
$2x^3+3x^2-12x+7 \le 0$
$<=>(2x+7)(x-1)^2 \le 0 => \left\{\begin{matrix}x=1\\y=-1\end{matrix}\right.$
 
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