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[nbngoc95]

GHPT: $\left\{\begin{matrix}

\sqrt{x^2+2y+3}+2y-3=0\\ 

2(2y^3+x^3)+3y(x+1)^2+6x(x+1)+2=0

\end{matrix}\right.$

[minhdat881439]

GHPT: $\left\{\begin{matrix} \sqrt{x^2+2y+3}+2y-3=0\\ 2(2y^3+x^3)+3y(x+1)^2+6x(x+1)+2=0 \end{matrix}\right.$

 

PT2$\Leftrightarrow 4y^3+2(x^3+1)+3y(x+1)^2+6x(x+1)=0$ 

$\Leftrightarrow 4y^3+2(x+1)(x^2-x+1)+3y(x+1)^2+6x(x+1)=0$

$\Leftrightarrow 4y^3+2(x+1)^3+3y(x+1)^2=0$ 

$\Leftrightarrow 4+2t^3+3t^2=0$ với $t=\dfrac{x+1}{y} \Rightarrow \dfrac{x+1}{y}=-2$

Thay vào PT(1) ta có $\sqrt{x^2-x+2}=x+4$

ĐK $x\geq -4$

Hệ có nghiệm $(x;y)=\left(\dfrac{-14}{19};\dfrac{5}{18} \right)$