ai giúp em với ạ em cần gấp $\left \{ {8{x^{3} - 12x^{2} + 8x = y^{3} + y + 2} \atop {x^{2} + y^{2} = 2}} \right.$

Xét pt $8x^3  - 12x^2+8x=y^3+y+2$

$↔ (8x^3 - 12x^2 + 6x - 1) + 2x+1-y^3-y-2=0\\↔ (2x-1)^3 - y^3 + 2x-y-1=0\\↔ (2x-1-y)[(2x-1)^2 + (2x-1)y+y^2] +(2x-y-1)=0\\↔ (2x-y-1) [(2x-1)^2 + (2x-1)y+y^2+1)]=0(1)$

Do $(2x-1)^2 +(2x-1)y+y^2+1=0$

$= (2x-1)^2+2.(2x-1) . \dfrac{1}{2}y+(\dfrac{1}{2}y)^2+\dfrac{3}{4}y^2+1\\=(2x-1+\dfrac{1}{2}y)^2 +\dfrac{3}{4}y^2+1\ge 1>0∀x,y\\(1)↔ 2x-y-1=0\\↔ y=2x-1$

Xét pt $x^2+y^2=2$

$↔ x^2+4x^2-4x+1=2\\↔ 5x^2-4x-1=0\\↔ (x-1)(5x+1)=0\\↔\left[ \begin{array}{l}x=1\\x=\dfrac{-1}{5}\end{array} \right.$

Với $x=1\to y=1$

Với $x=\dfrac{-1}{5}\to y=±\dfrac{7}{5}$

Thử lại ta thấy $(x;y)=(1;1), (\dfrac{-1}{5};\dfrac{-7}{5})$ thỏa mãn hpt.

Vậy hpt có nghiệm $(x;y)=(1;1), (\dfrac{-1}{5}; \dfrac{-7}{5})$